tạp chí physics for you tháng 2 năm 2016

A t P A = 8 Dual Nature of Matter and Radiation Atoms and Nuclei | By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second

Physics for you |FEBRUARY ‘16 7

Physics Musing Problem Set 31 8

JEE Main Practice Paper 10

Core Concept 22 JEE Workouts 26 JEE Accelerated Learning Series 31

Brain Map 46 Ace Your Way CBSE XII 54 AIPMT Practice Paper 63 Physics Musing Solution Set 30 73

Exam Prep 2016 75 You Ask We Answer 82

Live Physics 83

At a Glance 2015 84 Crossword 85

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Volume 24 No 2 February 2016

single oPtion correct tyPe

1 A particle starts from rest at A and moves with

uniform acceleration a m s–2 in a straight line After

1/a seconds, a second particle starts from A and

moves with uniform velocity u in the same line and

same direction If u > 2 m s–1 then during the entire

motion, the second particle remains ahead of first

particle for a duration

2 A block of mass 100 g moves with a speed

of 5 m s–1 at the highest point in a closed circular

tube of radius 10 cm kept in a vertical plane The

cross-section of the tube is such that the block just fits

in it The block makes several oscillations inside the

tube and finally stops at the lowest point The work

done by the tube on the block during the process is

3 Two identical balls A and B are released from the

positions shown in figure They collide elastically

on horizontal portion MN All surfaces are smooth

The ratio of heights attained by A and B after

collision will be (neglect energy loss at M and N)

4 A stone of mass m, tied to the end of a string, is

whirled around in a horizontal circle (Neglect the

force due to gravity.) The length of the string is

reduced gradually, keeping the angular momentum

of the stone about the centre of the circle constant

Then, the tension in the string is given by T = Ar n,

where A is a constant, r is the instantaneous radius

of the circle, and n is

5 Consider two hollow glass spheres, one containing water and the other containing mercury Each liquid fills about one-tenth of the volume of the sphere In zero gravity environment

(a) water and mercury float freely inside the spheres

(b) water forms a layer on the glass while mercury floats

(c) mercury forms a layer on the glass while water floats

(d) water and mercury both form a layer on the glass

subjective tyPe

acceleration of 4 m s–2 covers half of its total path during the last second of its motion Find the time taken and the total distance covered

7 A cone of height h and base radius r is fixed base to base on a hemisphere of equal radius Find h so that

the centre of gravity of the composite solid lies on the common base (Assume same density for both objects.)

8 A rod PQ of length l is pivoted at an end P and

freely rotated in a horizontal plane at an angular

speed w about a vertical axis passing through P If

coefficient of linear expansion of material of rod is

a, find the percentage change in its angular velocity ,

if temperature of system is increased by DT.

9 Distance between the centres of two stars is 10a The masses of these stars are M and 16 M and their radii

a and 2a, respectively A body of mass m is fired

straight from the surface of the large star towards the smaller star What should be its minimum initial speed to reach the surface of the smaller star?

10 The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz The fundamental frequency of the closed organ pipe is 110 Hz Find the lengths of the pipes (Take, speed of sound in air = 330 m s–1)

PaPer-i (single oPtion correct tyPe)

1 From a certain height, two bodies are projected

horizontally with velocities 5 m s–1 and 15 m s–1

They reach the ground in time t1 and t2 respectively

Then

u/2 respectively Both the balls cover the same

horizontal distance before returning to the plane of

projection If the angle of projection of ball B is 15°

with the horizontal, then the angle of projection of

A is

(a) (1/2) sin–1 (1/8) (b) (1/4) sin–1 (1/8)

(c) (1/3) sin–1 (1/8) (d) sin–1 (1/8)

3 A projectile is fired at an angle 30° to the horizontal

such that the vertical component of its initial

velocity is 80 m s–1 Find approximate velocity of

the projectile at time T/4 where T is time of flight.

4 A particle is projected with a velocity u so that

the horizontal range is twice the greatest height

attained, then the greatest height attained is

5 A bomb is dropped from a plane flying horizontally

with velocity 720 km h–1 at an altitude of 980 m

The bomb will hit the ground after (Take g = 9.8 m s–2)

6 The trajectory equation of a particle is given as

y = 4x – 2x2 where x and y are the horizontal and

vertical displacements of the particle in m from origin(point of projection) Find the maximum

distance of the projectile from x-axis.

(Take g = 9.8 m s–2)

horizontal distance, the angle at which the ball may

be thrown with a speed of 56 m s–1 without hitting

the ceiling of the hall is (Take g = 9.8 m s–2)

9 Two paper screens A and B are separated by 150 m

A bullet pierces A and then B The hole in B is 15 cm below the hole in A If the bullet is travelling horizontally at the time of hitting A, then the velocity of the bullet at A is (Take g = 10 m s–2)(a) 500 3 m s−1 (b) 200 3 m s−1(c) 100 3 m s−1 (d) 300 3 m s−1

10 A flag is mounted on a car moving due north with velocity of 20 km h–1 Strong winds are blowing due west with velocity of 20 km h–1 The flag will point

Motion in a Plane

Contributed by : Shiv R Goel, Intelli Quest, 9878359179

(a) At time t B, both trains have the same velocity.

(b) Both trains have the same velocity at some time

after t B

(c) Both trains have the same velocity at some time

before t B

(d) Nowhere the trains have some velocity

12 A wedge is placed on a smooth horizontal plain

and a rat runs on its sloping side The velocity of

wedge is v = 4 m s–1 towards right What should be

the velocity of rat with respect to wedge (u), so that

the rat appear to move in vertical direction to an

observer standing on ground?

13 A plank is moving on ground with a velocity v and a

block is moving on the plank with respect to it with

a velocity u as shown in figure What is the velocity

of block with respect to ground?

(a) v–u towards right (b) v–u towards left

(c) u towards right (d) none of these

14 A man is crossing a river flowing with velocity

of 5 m s–1 He reaches a point directly across at a

distance of 60 m in 5 s His velocity in still water

should be

15 Two particles P1 and P2 are moving with velocities

v1 and v2 respectively Which of the statements

about their relative velocity v r is true?

(a) v r cannot be greater than v1 + v2(b) v r cannot be greater than v1 – v2(c) v r > (v1 + v2 )

(d) v r < (v1 – v2 )

16 A boat having a speed of 5 km h–1 in still water, crosses a river of width 1 km along the shortest possible path in 15 minutes The speed of the river

in km h–1 is

17 A man can swim in still water with a speed of 2 m s–1 If

he wants to cross a river of water current of speed

3 m s–1 along shortest possible path, then in which direction should he swim?

(a) At an angle 120° to the water current(b) At an angle 150° to the water current(c) At an angle 90° to the water current(d) None of these

18 A ship is travelling due east at 10 km h–1 What must be the speed of a second ship heading 30° east

of north if it is always due north of the first ship?

in water to reach the other bank by shortest route

20 A bird flies to and fro between two cars which move

with velocities v1 = 20 m s–1 and v2 = 30 m s–1 If

the speed of the bird is v3 = 10 m s–1 and the initial

distance of separation between them is d = 2 km,

find the total distance covered by the bird till the cars meet

(a) 2000 m (b) 1000 m

21 A car is moving towards a wall with a fixed velocity

of 20 m s–1 When its distance from the wall is 100 m,

a bee starts to move towards the jeep with a constant

velocity 5 m s–1 The time taken by bee to reach the

jeep is

22 A stone is allowed to fall from the top of a tower and

covers half of the height of tower in the last second

of the journey The time taken by the stone to reach

the foot of the tower is

(c) (2+ 2) s (d) (2± 2) s

23 If a b c  , , are unit vectors such that a b c  + − = 0,

then the angle between a and b is

(a) π

π3(c) π

23

π

24 The sum of magnitudes of two forces acting at a

point is 16 N and magnitude of their resultant is

8 3 N If the resultant is at 90° with the force of

smaller magnitude, then their magnitudes(in N)

are

25 The resultant of two forces acting at an angle of 150°

is 10 N and is perpendicular to one of the forces

The other force is

PaPer-ii (one or More oPtions correct tyPe)

26 The velocity time graph of two bodies A and B is

given here Choose correct statements

(a) Acceleration of B > acceleration of A.

(b) Acceleration of A > acceleration of B.

(c) Both are starting from same point

(d) A covers greater distance than B in the same time.

27 A man wishes to throw two darts one by one at the

target at T so that they arrive at T at the same time

as shown in figure Mark the correct statements about the two projections

(a) Projectile that travels along trajectory A was

projected earlier

(b) Projectile that travels along trajectory B was

projected earlier

(c) Both were projected at same time

(d) Darts must be projected such that qA + qB = 90°

28 In the figure shown, two boats start simultaneously with different speeds relative to water Water flow speed is same for both the boats Mark the correct statements qA and qB are angles from y-axis at

which boats are heading at initial moment

(a) If v A > v B then for reaching the other bank simultaneously qA > qB

(b) In option (a), drift of boat A greater than boat B (c) If v B > v A and qA > qB , boat B reaches other bank earlier than boat A.

(d) If v B = v A and qA > qB , drift of A is greater.

29 A particle has an initial velocity of 4 4i+ j m s −1and an acceleration of −0 4 i m s–2, at what time will its speed be 5 m s–1?

30 A boat is traveling due east at 12 m s–1 A flag on the boat flaps at 53° north of west Another flag on the shore flaps due north

(a) Speed of wind with respect to ground is 16 m s–1(b) Speed of wind with respect to ground is 20 m s–1(c) Speed of wind with respect to boat is 20 m s–1(d) Speed of wind with respect to boat is 16 m s–1

Physics For you |FEBRUARY ‘16 15

PaPer-i (single oPtion correct tyPe)

31 A child is sliding down a slide in a playground with

a constant speed

Statement-1 : His kinetic energy is constant

Statement-2 : His mechanical energy is constant

(a) Statement-1 is true, statement-2 is true

and statement-2 is correct explanation for

statement-1

(b) Statement-1 is true, statement-2 is true and

statement-2 is not the correct explanation for

statement 1

(c) Statement-1 is true, statement-2 is false

(d) Statement-1 is false, statement-2 is true

32 Statement - 1 : Force F1 required to just lift block A

of mass m in case (i) is more than that in case (ii).

Statement -2 : Less work has to be done in case (ii)

to lift the block from rest to rest by a distance h.

F1 F2

(a) Statement-1 is true, statement-2 is true

and statement-2 is correct explanation for

statement-1

(b) Statement-1 is true, statement-2 is true and

statement-2 is not the correct explanation for

statement 1

(c) Statement-1 is true, statement-2 is false

(d) Statement-1 is false, statement-2 is true

33 A particle initially at rest is displaced from x = –10 m to

x = +10 m under the influence of force F as shown

in the figure Now the kinetic energy vs position

graph of the particle is

(a)

(b)

(c)

(d)

34 The potential energy of an object is given by

U(x) = 3x2 – 2x3, where U is in joules and x is in

meters

(a) x = 0 is stable and x = 1 is unstable.

(b) x = 0 is unstable and x = 1 is stable.

(c) x = 0 is stable and x = 1 is stable.

(d) x = 0 is unstable and x = 1 is unstable.

35 A boy blowing a whistle sends out air at one gram per second with a speed of 200 m s–1 Find his lung power

(a) 20 W (b) 0.2 W (c) 2 W (d) 200 W

36 A stone tied to a string of length 2 m is whirled in

a vertical circle with the other end of the string at the centre At a certain instant of time, the stone is

at its lowest position and has a speed 10 m s–1 The magnitude of the change in its velocity as it reaches

a position where the string is horizontal, is

37 A ball of mass m is hung on a thread The thread

is held taut and horizontal, and the ball is released

as shown At what angle between the thread and vertical will the tension in thread be equal to weight

in magnitude?

laws oF Motion; work, energy and Power

(a) 30° (b) cos−123

(c) cos−113  (d) never

38 A ball whose size is slightly smaller than width of the

tube of radius 2.5 m is projected from bottommost

point of a smooth tube fixed in a vertical plane with

velocity of 10 m s–1 If N1 and N2 are the normal

reactions exerted by inner side and outer side of the

tube on the ball

(a) N1> 0 for motion in ABC, N2 > 0 for motion in

39 A catapult on a level field tosses a 3 kg stone, a

horizontal distance of 100 m At second time, 3 kg

stone tossed in an identical fashion breaks apart in

the air into 2 pieces, one with a mass of 1 kg and

one with a mass of 2 kg Both of the pieces hit the

ground at the same time If the 1 kg piece lands a

distance of 180 m away from the catapult, how far

away from the catapult does the 2 kg piece land?

Ignore air resistance

40 A small particle of mass m is at rest on a horizontal

circular platform that is free to rotate about a

vertical axis through its center The particle is

located at a radius r from the axis, as shown in the

figure below The platform begins to rotate with

constant angular acceleration a Because of friction

between the particle and the platform, the particle

remains at rest with respect to the platform When

the platform has reached angular speed w, the

angle q between the static frictional force f s and the

inward radial direction is

r fs

(a) The magnitude of acceleration is constant.(b) The acceleration vector is along the tangent to the circle

(c) The velocity vector points along tangent to the circle

(d) The velocity and acceleration vectors are always perpendicular to each other

42 A 50 kg boy runs at a speed of 10 m s–1 and jumps onto a cart as shown in the figure The cart is initially

at rest If the speed of the cart with the boy on it is 2.50 m s–1, what is the mass of the cart?

43 In a one-dimensional collision, a particle of mass

2m collides with a particle of mass m at rest If the

particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

45 A spaceship of speed v0 travelling along +y axis

suddenly shots out one fourth of its part with speed

2v0 along +x-axis xy axes are fixed with respect to

ground The velocity of the remaining part is(a) 2

Physics For you |FEBRUARY ‘16 17

47 On a frictionless surface, a ball of mass M moving

at speed v collides elastically with another ball of

the same mass that is initially at rest After the

collision, the first ball moves at an angle q to its

initial direction and has a speed v/2 The second

ball's speed after the collision is

(a) 3

2cosq

48 A particle of mass m is moving along the x-axis with

speed v when it collides with a particle of mass 2m

initially at rest After the collision, the first particle

has come to rest, and the second particle has split

into two equal mass pieces that move at equal angle

q = 30° with the x-axis, as shown in the figure

below Which of the following statements correctly

describes the speeds of the two pieces?

(a) Each piece moves with speed v.

(b) One of the pieces moves with speed v, the other

moves with speed less than v.

(c) One of the pieces moves with speed v/2, the

other moves with speed greater than v/2.

(d) Each piece moves with speed greater than v/2.

49 Two objects of different mass and with same initial

speed, moving in a horizontal plane, collide head

on and move together at half their initial speed after

the collision Ratio of their masses is

50 A man of mass 3 M stands at one end of a plank of

length L which lies at rest on a frictionless surface

The man walks to other end of the plank If the

mass of the plank is M, then the distance that the

man moves relative to ground is

51 Block A, with a mass of 4 kg, is moving with a speed

of 2.0 m s–1 while block B, with a mass of 8 kg, is

moving in the opposite direction with a speed of

3 m s–1 The center of mass of the two block system

is moving with a velocity of (a) 1.3 3 m s–1 in the same direction as A

(b) 1.3 3 m s–1 in the same direction as B

(c) 2.7 3 m s–1 in the same direction as A

(d) 1.0 3 m s–1 in the same direction as B.

PaPer-ii (one or More oPtions correct tyPe)

52 You lift a suitcase from the floor and keep it on a table The work done by you on the suitcase does not depend on

(a) the path taken by the suitcase(b) the time taken by you in doing so(c) the weight of the suitcase

(d) your weight

53 Figure (a) shows a frictionless roller coaster track Figures (i), (ii), (iii) show potential energy and total energy for a car on roller coaster Figures (i), (ii), (iii) show drawing of actual roller coaster track because gravitational potential energy is directly

proportional to height, it is also U vs x graph.

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U

EtotalD

(a) In case (i), the car can negotiate the hill at C and

reach D.

(b) In case (ii), the motion is confined between

two turning points where the total energy and

potential energy curves intersect

(c) In case (iii), motion of car will be confined in

first valley between A and B.

(d) The turning points are attained where Etotal = U.

54 An object of mass 3m, initially at rest on a frictionless

horizontal surface, explodes breaking into two

fragments of mass m and 2m respectively Which

one of the following statements after the explosion

is true?

(a) Velocity of center of mass increases

(b) Speed of smaller fragment will be twice that of

larger fragment

(c) Fragments have equal magnitude of momentum

in ground frame but different magnitude of

momentum in center of mass frame

(d) Kinetic energy of system increases

55 Bodies of mass 0.5 kg, resting on a horizontal

frictionless tabletop, are connected with an

unstretched spring of length L = 20 cm, and of

spring constant 16 N m–1 The mass of the spring

is negligible At a certain moment the bodies are

given an initial speed of v0 = 0.36 m s–1, towards the wall on the right The body at the right collides with the wall totally elastically

(a) There will be 2 collisions with the wall

(b) After 1st collision, centre of mass comes to rest.(c) After 2nd collision, centre of mass moves to left

with speed v0.(d) After all collisions are over, the system oscillates about the centre of mass

u g

1

Physics For you |FEBRUARY ‘16 19

Here we have to look for velocity of wind with

respect to car So,

v w c/ =v w− = −v c 20 20i− j

This is in south-west direction

11 (c) : At some time before t B , slope of B will be

equal to slope of A Acceleration of A is zero always

whereas that of B is not zero.

15 (a) : v r lies between (v1 – v2) and (v1 + v2) depending

upon angle between v1 and v2

Hence at an angle of 150° to the water current

18 (d) : Speed of both boats towards east should be same

t = ±2 2 s

But t cannot be 2− 2seconds since it is less than

1 second which is not possible

27 (b) : Since maximum height attained by B is more,

so it will take more time to reach the target Hence

it should be projected earlier so that both reach

simultaneously For same range, sum of projection

angles is 90° if speed of projection is same, but here

speeds may be different

28 (a, b, c, d) : For reaching the other bank

simultaneously, their velocities along y direction

should be same

So, v A cos qA = v B cos qB , if v A > v B then cos qA < cos qB,

⇒ qA > qB Hence (a) is correct

Drift: x = (v sin q + u)t

For option (a): v A sin qA > v B sin qB, hence drift of

A is greater than B So (b) is correct.

Same will be true for option (d), hence (d) is also

correct

For option (c): v B cos qB > v A cos qA , so boat B

reaches earlier than A Hence (c) is correct.

29 (a, b) : Since acceleration is in x direction only,

velocity in y-direction will not change.

52 = v x2 + v y2 = v x2 + 42 ⇒ v x = ± 3 m s–1

v x = u x + a x t ⇒ t v u

a

x x x

Actual velocity of wind is towards north, because a

flag on shore flaps towards north We have  

Physics For you |FEBRUARY ‘16 21

40 (d) : tanq a tan

w

aw

a

R R

t

1 2

1 Anu Sharma (Delhi)

2 Shruti Gupta (Delhi)

3 Rohan Kashyap (Haryana)

4 Atriz Roy (WB)

Solution Senders December 2015

1 Sakchhi Kumari (Jharkhand)

2 Puneet Goel (UP)

solution oF January 2016 crossword

In this segment, we would learn and gradually master

solving capacitive circuit questions

To begin with, let me start asking two basic questions:

1 What is meant by series combination?

2 What is a parallel combination?

Maximum book says that in a series combination the

charges on all capacitors are identical But is it really

Considering, initially all capacitors are uncharged There

was no potential difference across the combination

before closing the switch, but on connecting the battery

a potential difference equal to emf of the cell has to be

maintained across the combination which can come only

if electric field is set up between the plates of capacitor

What the battery does is, it pulls out electrons from

one plate of capacitor and deposits it to the other plate

of the other extreme end's capacitor and thereafter by

induction and conduction charges are induced on other

plates and capacitors

Consider any intermediate capacitors, say C1 and C2

From left plate of C1

C1

2 –Q

isolated

charge –Q was pulled

by battery due to which

+Q appeared on it By

induction –Q appears on

right plate of C1 but right plate of C1 and left plate of

C2 forms an isolated part of circuit By saying isolated

we mean, they are not physically connected to others,

hence the net charge on them has to remain conserved

and hence since before charging the capacitors were uncharged, the summation of charges on these plates

has to be zero i.e.

Sqjunction = 0 [Kirchhoff's junction law]

from the left face of C2 due to which +Q appears on

it (one gains, other loses by same amount)

But what if the capacitors were already charged before connecting them to the battery?

Then the Sqjunction ≠ 0 and hence the charges on the capacitors would not be equal even if they were in series combination

Just remember one line - If between any two terminals,

say A and B, there exists only one path which leads from terminals A to B, the capacitors are said to be in series but use series combination formula i.e.

Supposedly now, none

of the capacitors were initially charged, what is the potential difference across each capacitor?

the ratio of capacitances For example if both C1 and

C2 are doubled, the potential difference across both remains unchanged

Capacitive Circuits

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

physics for you |february ‘16 23

Now let me answer the 2nd question raised at the

beginning

We say two or more capacitors to be in parallel if their

ends are maintained at common potential They need

not be placed geometrically parallel to each other

For example,

C1

C2

C3B

A

Notice that left plates of all the three capacitors are

connected to terminal A whereas their right plates to

terminal B through connecting conducting wires and

since all the points of a conductor are equipotential at

electrostatic condition, we conclude that a common

potential difference is maintained across each, hence

equivalent capacitance,

Ceq = C1 + C2 + C3

With these basic understanding being cleared, now let

us move to our next segment where we learn how to

solve complicated circuits involving multiple batteries

and capacitors

Our approach will be slightly different from what is

followed by most books

Just follow these basic steps:

1 If none of the point in the given circuit is grounded,

we can choose any one arbitrary point in the circuit

to be at zero potential Depending on the choice

of zero potential point, the potential of all other

points will be dependent but the potential difference

between any two points would not change and for

any capacitor it is the potential difference which

is important and not the potential of each of its

terminals

I prefer taking zero volt to be at the lower potential

terminal of the largest e.m.f's cell connected This

simplifies calculation

2 Distribute the potential of all points (junctions) in

the circuit, either in terms of known or unknown

variables, by using the fact that

-(a) All points of conducting wire are at same potential

(b) The potential difference across the cell is equal

to its emf (at electrostatic condition)

For example,

10 V ( )x( + 10)x

(c) If the charge on a capacitor is not known, the potential difference across it would not be known, hence new variables would be required for it

3 Count the number of variables taken, since we need

to frame exactly same number of equations by using

Kirchhoff's junction law, i.e.

4 Solve simultaneous equations obtained

Now, let us apply whatever we learnt

Q.1 : A capacitor of capacitance 5 mF is charged with an initial charge of 50 mC and then connected to another uncharged capacitor of 20 mF with a battery of emf

Soln.: After closing the switch, the potential of different

points has been shown as below:

Q.2 : The shown network

is a small segment of

a large circuit and the

potential of the three

terminals are marked

Find the potential

V1

C1

C3

Soln.: Here, since the

potential of certain points

are already marked, the

choice of zero potential

is not our own

Let the junction be at

Q.3 : In the given circuit, after steady state, charge on

3 mF is found to be 120 mC then find

(i) emf of cell

(ii) potential difference between points x and y.

2 F  4 F  ( )y

4 F  ( )x

This is also the potential difference across a parallel

combination of 2 mF, 3 mF and 4 mF Hence,

From the diagram,9

Sol.: Clearly 6 mF and 3 mF are in series, hence we

can replace them with 6 3

×

potential of all points as below:

10 V 2 F

20 V

4 F  ( + 10)x ( )x (20)

physics for you |february ‘16 25

Q.6 : Find (i) charge flown through the switch,

(ii) work done by both the batteries after closing the

switch in the circuit below

The result shows both the capacitors have identical

charge and that should have been since they are in

series

So alternatively, the charge could have been found out imagining a single capacitor and single cell in the loop

\ q3 = q6 = Cloop eloop =

×+



3 63 6 ×(20 10 20− )= CmHence, the higher potential terminal of both the capacitors will have +20 mC while the other has –20 mC charge

40

3

–20  C +20  C

After closing the switch:

(20)

(30)

Clearly, potential difference across 3 mF and 6 mF are

10 V and 20 V respectively now

q6 = 6 × 20 = 120 mCThe right plate of 6 mF capacitor initially had –20 mC but now has –120 mC, hence 100 mC charge flows through the wire connected to it and hence through

20 V battery also

= 100 × 20 = 2000 mJ = 2 mJSimilarly, charge flown through 10 V battery = 50 mC

\ Work done by 10 V battery = 50 × 10

In a certain region of space, there exists a uniform

and constant electric field of magnitude E along the

positive y-axis of a co-ordinate system A charged

particle of mass m and charge –q (q > 0) is projected

from the origin with speed 2v at an angle of 60° with

the positive x-axis in x–y plane When the x-coordinate

of particle becomes 3mv qE2/ , a uniform and constant

magnetic field of strength B is also switched on along

2 x-coordinate of the particle as a function of time

after the magnetic field is switched on is

qE

mv qB

3 z-coordinate of the particle as a function of time

after the magnetic field is switched on is

Paragraph-2

There is a uniformly charged

ring having radius R An

infinite line charge (charge per unit length l) is placed along a diameter of the ring (in gravity free space) Total charge on the

ring Q= 4 2l An electron R

of mass m is released from

rest on the axis of the ring at

a distance x= 3 from the R

Physics for you |February ‘16 27

6 Potential difference between points A(x = 3R)

34

(d) none of these

Paragraph-3

Pulfrich refractometer is used to

measure the refractive index

of solids and liquid It consists

of right angled prism A having

its two faces perfectly plane

One of the face is horizontal

and the other is vertical as

shown in figure The solid B whose refractive index

is to be determined is taken having two faces cut

perpendicular to one another Light is incident in a

direction parallel to the horizontal surface so that the

light entering the prism A is at critical angle C Finally, it

emerges from the prism at an angle i Let the refractive

index of the solid be m and that of the prism A be m0

(which is known) Here m0 > m and by measuring i, m

can be determined

7 Refractive index of the solid (m) in terms of m0 and i is

(a) m02+ sin i 2 (b) m0 + sin2i

(c) m02− sin i 2 2 (d) m02− sin i2

8 If m0 = 2 and the ray just fails to emerge from the

prism, refractive index m of the solid will be

9 A ray of light is incident normally

on the horizontal face of the slab

and just fails to emerge from the

diagonal face of the prism If prism

angle is 30°, refractive index of the

10 Calculate the number of photoelectrons emited per second

11 It is observed that photoelectron emission stops at

a certain time t after the light source is switched on

It is due to the retarding potential developed in the metallic sphere due to left over positive charges The

stopping potential (V) can be represented as

12 Evaluate time t, mentioned in question 11.

Paragraph-5

In figure shown, the rod has a resistance R, the horizontal

rails have negligible friction A battery of e.m.f e and negligible internal resistance is connected between points

a and b The rod is initially at rest.

13 The velocity of the rod as a function of time t (where t = mR/B2l2) is

Physics for you |February ‘16 29

15 The net current through the circuit when the rod

attains its terminal speed is

1 (a) : At first, particle will travel along parabolic path

m y

0 =−

mv qE

y= y+ y 0=2 sin60° − 3 =0

Hence at point A, velocity will be purely along

x-axis and it will be 2vcos60° = v.

2 (b) : Now magnetic field is switched on along y-axis

Now its path will be helical as shown below with

increasing pitch towards negative y-axis.

4 (a) : Electric field at distance x,

lπe2

13

2 2 38

lπe2

7 (d) : sinC= m ⇒sin[ ° − =r]

m

mm

8 (c) : If ray just fails to emerge, i = 90°

If E is the energy of the single photon and h the

efficiency of the photon to liberate an electron, the

number of ejected electrons is

11 (b) : The emission of electrons from a metallic sphere

leaves it positively charged As the potential of the

charged sphere begins to rise, it attracts emitted

electrons The emission of electrons will stop when

the kinetic energy of the electrons is neutralised by

the retarding potential of the sphere So, we have

4

14

13 (a) : The current due to the battery at any instant, I = e/R

The magnetic force due to this current

R

B= = eThis magnetic force will accelerate the rod from its position of rest The motional e.m.f developed in

the rod is Blv.

R

induced=The magnetic force due to the induced current

R

induced= 2 2

From Fleming’s left hand rule, force F B is to the

right and Finduced is to the left

Net force on the rod = F B – Finduced.From Newton’s second law,

On separating variables and integrating speed from

0 to v and time from 0 to t, we have

dv Bvl

Bl

dv Bvl

14 (c) : The rod will attain a terminal velocity at t → ∞

i.e., when e –t/t = 0, the velocity is independent of time

v Bl

T = e

15 (d) : The induced current Iinduced = Blv/R When the

rod has attained terminal speed,

Physics for you |FEBRUARY ‘16 31

dual nature of matter and radiation

The wave nature of light was established by Maxwell’s

equations of electromagnetism and Hertz experiment

during the generation and detection of electromagnetic

waves

The discoveries of photoelectric effect by Hertz,

Compton effect by Compton, Stark effect by Stark were

explained by Planck’s quantum theory of light According

to which, the light consists of the packets of energy

which travel in straight line, with the speed of light

Each packet of energy is called photon or quantum of light

Energy of each photon is E = hu = hc/l where h is Planck’s

constant, u is the frequency of light, c is the velocity of

light and l is the wavelength of light This, established

the particle nature of light As some phenomena of

light were explained by wave theory of light and some

by particle nature of light, hence it was concluded that

light is of dual nature

Photons

A photon is massless i.e., zero rest mass and moves with

the velocity of light in vacuum i.e., c = 3 × 108 m s–1

It can never be brought to rest

Photons also carry momentum p.

per unit area per unit time or power carried per unit area

A t

P A

=

8 Dual Nature of Matter and Radiation Atoms and Nuclei |

By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy Thus, photon energy is independent

or a new photon may be created

Number of photons emitted per second of frequency u

from a lamp of power P is

n P h

P hc

ul

•

phenomenon of emission of electrons from the

surface of a metal under the application of a strong

electric field

Secondary emission : It is the phenomenon of

•

emission of electrons from the surface of metal in

large number when fast moving electrons called

primary electrons strike the metal surface

KEYPOINT

Photons are not deflected by electric and magnetic

•

fields, which shows that they are neutral and do

not carry any charge

The energy of photon depends upon the frequency

It is the phenomenon in which electrons are emitted

from a metal surface when radiation of sufficient energy

falls on it

hertz’s observations

The phenomenon of photoelectric emission was

discovered in 1887 by Heinrich Hertz during his

electromagnetic wave experiment Hertz found that

high voltage sparks across detector loop were enhanced

when an emitter plate was illuminated by ultraviolet

light from an arc lamp When the emitter plate was

illuminated by ultraviolet light, some electrons near the

surface of the metallic emitter plate absorb energy from

the ultraviolet rays This enables them to overcome the

force of attraction due to positive ions in the material of

the emitter and finally escape from the emitter surface

into the surrounding space, hence enhancing the high

voltage sparks across the detector

hallwachs’ and lenard’s observations

When ultraviolet radiation was allowed to fall on the

emitter plate of an evacuated glass tube enclosing two

electrodes, current flowed in the circuit

After the discovery of electrons, it became evident that

the incident light causes electrons to be emitted from

the emitter plate It was also observed that no electrons

were emitted at all when the frequency of the incident

light was smaller than a certain minimum value This

minimum frequency is known as threshold frequency

and depends on the nature of the material of the emitter

plate Substances which emit electrons when illuminated

by light are known as photosensitive substances and the

emitted electrons are known as photoelectrons

For a given metal, there exits a certain minimum frequency of light radiation below which no photoelectric emission takes place This minimum frequency of radiation is known as threshold frequency (u0)

The minimum energy of incident radiation needed to eject the electrons from metal surface is known as work

function (f0) of that surface Work function is related to

ejected photoelectron and u is the frequency of incident light photon

As f0 = hu0, hence Einstein’s equation may be written as

h(u u− 0)=Kmax=1mvmax2

2

If in a photoelectric tube we apply a negative potential, then for a certain minimum negative potential the photoelectric current becomes zero This negative

potential is known as stopping potential (V0) It is, thus, a

measure of maximum kinetic energy of photoelectrons,

Physics for you |FEBRUARY ‘16 33

Wave nature of matter

Radiation has dual nature, wave and particle nature

The nature of experiment determines whether a wave

or a particle description is best suited for understanding

the experimental result Reasoning that radiation and

matter should be symmetrical in nature, Louis Victor

de Broglie attributed a wave-like character to matter

The waves associated with the moving material particles

are called matter waves or de Broglie waves

where p is the momentum of the particle and K is the

kinetic energy of the particle and m is mass of the

particle

de Broglie wavelength is independent of the charge

•

and nature of the material particle

If the rest mass of a particle is

de-Broglie wavelength of electron decreases

davisson and Germer experiment

Davisson and Germer experiment proves the

•

concept of wave nature of matter particles In a

crystal lattice, the interatomic distance between the

layers and de-Broglie wavelengths of an electron

are nearly of same order So, diffraction of electron

beam can be observed through crystals This

experiment uses an electron gun to produce fine

beam of electrons which can be accelerated to any

desired velocity by applying suitable voltage across the gun

A fine beam of electrons is made to fall on the surface

•

of nickel crystal The electrons are scattered in all directions by the atoms of the crystal The intensity

of the electron beam, scattered in a given direction,

is measured by the electron detector, which can be rotated, on a circular scale

Incident beam

Electron gun

Detector

Scatbeamtered

Ni-crystal Experimental set up

of the intensity I of the scattered electrons with

the angles of scattering f at different accelerating voltages It is found that intensity is different for different angles of scattering Further, the maximum intensity is obtained due to constructive interference

of electrons scattered from different layers of regularly spaced atoms of the crystals

It is found that angle q between the scattered beam

of electrons with the plane of atoms of crystal, when scattering angle f = 50° is

q + f + q = 180°

2q = 180° – 50° or q = 65°

Now using Bragg’s law, 2d sin q = nl

but for first order diffraction, l = 2d sin q

where d = 0.91 Å is distance between two successive

layers of atoms in Ni crystal

or l = 2 × 0.91 sin 65° or l = 1.66 Å

This is the value of wavelength of electron as measured

by Davission and Germer experiment However, the

de-Broglie wavelength of an electron accelerated

through potential difference V = 54 volts is

As the two results are same, so this experiment

proves the wave nature of electron and hence of a

particle in general

electron microscope

The wave nature of electrons affords us the possibility of

having probes of very short wavelength Electrons speed

up to high energies, using an accelerating voltage of, say,

50 kV have a de Broglie wavelength of 0.0055 nm, This

is about 105 times smaller than that of visible light

An electron microscope is a device that exploits the

wave nature of electrons Theoretically, the resolving

limit of the electron microscope, using electrons of

50 keV, would be 0.0055 nm

However, in practice, the electron beam needs to be

focussed using electric and magnetic fields as lenses

(much like a beam of light is focussed using optical lenses)

These limit the resolution to about 0.2 nm, which is still

1000 times better than that of the optical microscopes

The electron microscope, with its high magnifying and

resolving powers, is one of the most indispensable and

powerful tools for research in science, medicine and industry

SELFCHECK

1 Match List I (Fundamental Experiment) with List-II

(its conclusion) and select the correct option from

the choices given below the list :

(a) P - (ii), Q - (i), R - (iii)(b) P - (iv), Q - (iii), R - (ii)(c) P - (i), Q - (iv), R - (iii)(d) P - (ii), Q - (iv), R - (iii) (JEE Main 2015)

2 de-Broglie wavelength of an electron accelerated

by a voltage of 50 V is close to (|e| = 1.6 × 10–19 C,

is gradually changed The plate current I of the

photocell varies as follows(a)

atoms and nuclei

alpha-particle scattering experiment

At the suggestions of Rutherford in 1911 Geiger and Marsden performed a-particle scattering experiment

Gold foil

Screen

Source of -particles 

Geiger-Marsden scattering experiment.

e entire apparatus is placed

in a vacuum chamber

They directed a beam of 5.5 MeV a-particles emitted from a 214 83 Bi radioactive source at a thin metal foil made of gold The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 10–7 m Alpha particles emitted by radioactive source were collimated into a narrow beam by passing through lead bricks

Physics for you |FEBRUARY ‘16 35

The scattered a-particles were received by a rotatable

detector with zinc sulphide screen and a microscope

Distribution of the number of scattered particles was

studied as a function of angle of scattering by flashes

or scintillations produced by striking a-particles on the

zinc sulphide screen

Schematic arrangement of the Geiger-Marsden

Detector Backward scattering

of a very small fraction

ZnS screen

Beam of -particles

 Source of

without any deflection This shows that most of the

space in an atom is empty

• Few a-particles got scattered, deflecting at various

angles from 0 to p This shows that atom has a small

positively charged core called nucleus at centre of

atom, which deflects the positively charged a-particles

at different angles depending on their distance from

shows that size of nucleus

is very small, nearly 1/8000

times the size of atom

deflection of number of particles with angle of

deflection q

Impact parameter

velocity vector of the alpha-particle from the centre

of the nucleus

b

 nucleus

N( )

 = 180° 0°

The trajectory traced by an alpha particle depends

•

on its impact parameter b Rutherford had

analytically calculated the relation between the

impact parameter b and the scattering angle q,

• b = 0, then by above relation cot q/2 = 0 or

q/2 = 90° or q = 180° i.e., in case of head on collision,

the impact parameter is zero and the alpha-particle rebounds back

If

• b = ∞, then by above relation cot q/2 = ∞ or

q/2 = 0° or q = 0° i.e., the alpha particle goes nearly

undeviated for a large impact parameter

distance of closest approach : estimation

of nuclear size

Suppose an

v moves directly towards the centre of the nucleus

of an atom As it approaches the positive nucleus,

it experiences Coulombic repulsion and its kinetic energy gets progressively converted into electrostatic

potential energy At a certain distance r0 from the nucleus, the a-particle stops for a moment and then

begin to retrace its path The distance r0 is called the distance of closest approach

Nucleus

+

+ +

+ +

Distance of closest approach

Hence radius of nucleus must be smaller than r0

models for structure of an atom

thomson’s model of atom

The atom as a whole is electrically neutral because

•

the positive charge present on the atom (sphere) is

equal to the negative charge of electrons present in

the sphere Atom is positively charged sphere of

radius 10–10 m in which electrons are embedded in

between The positive charge and the whole mass of

the atom is uniformly distributed throughout the

sphere

shortcomings of thomson model

The spectrum of atoms cannot be explained with

•

the help of this model

Scattering of

the help of this model

rutherford model of atom

On the basis of his study of the scattering of

•

a-particles, Rutherford postulated the following

model of the atom

Atom is a sphere of diameter about 10

of its positive charge and most of its mass is

concentrated in the central part called the nucleus

The diameter of the nucleus is of the order of

•

10–14 m

The space around the nucleus is virtually

•

empty with electrons revolving around the nucleus

in the same way as the planets revolve around the

sun

The electrostatic attraction of the nucleus provides

•

centripetal force to the orbiting electrons

Total positive charge in nucleus is equal to the total

•

negative charge of the orbiting electrons

rutherford scattering formula

The number of

by a target are given by

where, N0 = number of a-particles that strike the

unit area of the scatter

n = number of target atoms per unit volume

0 0

2

2

14

212

14

where K is kinetic energy of a-particle.

Bohr’s theory of hydrogen atom

An electron in an atom moves in a circular orbit about the nucleus under the influence of Coulomb’s force of attraction between the electron and nucleus As the atom as a whole is stable, the Coulombian force of attraction provides necessary centripetal force

e r

mv r

Equation (iii) shows that the radii of the permitted orbits

vary as the square of n, called the principal quantum number For the smallest orbit n = 1, substituting the values of h, e0, m and e we have radius of first orbit

r1 = 0.529 × 10–10 m = 0.529 ÅThis calculation shows that the atom is about 10–10 m

in diameter

Velocity of revolving electron

To obtain the velocity of the revolving electron, we

substitute the value of r from equation (iii) in (ii), we get

orbital frequency of electron

The orbital frequency of electron is given by,

•

ue

= me

n h

4 0

2 3 3

This expression shows that the orbital frequency of an

electron is inversely proportional to the cube of n.

Physics for you |FEBRUARY ‘16 37

4

2 2 0 2

SELFCHECK

4 The de-Broglie wavelength associated with the

electron in the n = 4 level is

(a) two times the de-Broglie wavelength of the

electron in the ground state

(b) four times the de-Broglie wavelength of the

electron in the ground state

(c) half of the de-Broglie wavelength of the electron

in the ground state

(d) 1/4th of the de-Broglie wavelength of the

electron in the ground state

(JEE Main 2015)

5 As an electron makes a transition from an excited state

to the ground state of a hydrogen -like atom/ion

(a) kinetic energy decreases, potential energy

increases but total energy remains same

(b) kinetic energy and total energy decrease but

potential energy increases

(c) its kinetic energy increases but potential energy

and total energy decrease

(d) kinetic energy, potential energy and total energy

decrease

(JEE Main 2015)

6 If one were to apply Bohr model to a particle of

mass m and charge q moving in a plane under the

influence of a magnetic field B, the energy of the

charged particle in the nth level will be

7 In a hydrogen like atom electron makes transition

from an energy level with quantum number n to

another with quantum number (n – 1) If n > > 1,

the frequency of radiation emitted is proportional to

spectral series of hydrogen atom

The wavelength of different lines of series can be found from the following relation

ul

lyman series

This series consists of wavelength which are emitted when

electron jumps from an outer orbit to the first orbit i.e., the electron jumping to K orbit gives rise to Lyman series Here, n1 = 1 and n2 = 2, 3, 4 ∞

The wavelength of different lines of Lyman series are

First line : In this case n1 = 1 and n2 = 2

34

89

Series limit : In this case, n1 = 1 and n2 = ∞

Balmer series

This series consists of all wavelengths which are emitted when an electron jumps from an outer orbit to the

second orbit i.e., the electron jumping to L orbit gives

rise to Balmer series

Here, n1 = 2 and n2 = 3, 4, 5 ∞

The wavelength of different lines of Balmer series are

First line : In this case n1 = 2 and n2 = 3,

536

316

This series consists of all wavelengths which are emitted

when an electron jumps from an outer orbit to the third

orbit i.e., the electron jumping to M orbit gives rise to

Paschen series

Here, n1 = 3 and n2 = 4, 5, 6 ∞

The different wavelengths of this series can be obtained

from the formula

For the first line, the wavelength is 18750 Å This series

lies in infra-red region

Brackett series

This series consists of all wavelengths which are emitted

when an electron jumps from an outer orbit to the

fourth orbit i.e., the electron jumping to N orbit gives

rise to Brackett series

Here, n1 = 4 and n2 = 5, 6, 7 ∞

The different wavelengths of this series can be obtained

from the formula

l=R( ) −n 

where, n2 = 6, 7, 8 ∞This series lies in far infrared region

ionisation and excitation energy

Ionisation energy of an atom is defined as the energy

required to ionise it i.e., to make the electron jump from its

present orbit to infinite orbit

Thus, ionisation energy of hydrogen atom in the ground

state = E∞ – E1 = 0 –(– 13.6 eV) = + 13.6 eVThe potential through which an electron is to be accelerated so that it acquires energy equal to the ionization energy is called the ionisation potential Therefore, ionisation potential of hydrogen atom in its ground state is 13.6 V

Excitation energy is the energy required to excite an electron from a lower energy level to a higher energy level Thus, first excitation energy of hydrogen atom

= E2 – E1 = – 3.4 – (– 13.6) eV = 10.2 eVSimilarly second excitation energy of hydrogen

an electron transition from n = 2 to n = 1 If the

wavelengths of emitted radiation are l1, l2, l3 and

l4 respectively then approximately which one of the following is correct?

(a) l1 = 2l2 = 3l3 = 4l4(b) 4l1 = 2l2 = 2l3 = l4(c) l1 = 2l2 = 2l3 = l4(d) l1 = l2 = 4l3 = 9l4 (JEE Main 2014)

9 The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons These electrons are made to enter

a magnetic field of 3 × 10–4 T If the radius of the

Physics for you |FEBRUARY ‘16 39

largest circular path followed by these electrons is

10.0 mm, the work function of the metal is close to

(JEE Main 2014)

nucleus

In every atom, the positive charge is concentrated at the

centre of the atom forming its nucleus The order of size

of nucleus is 10–15 m or fermi

The order of size of atom is 10–10 m or Å Only protons

and neutrons can exist inside the nucleus Electrons

orbit around the nucleus in certain orbits and are called

Three forces are interacting between nucleons

Gravitational force which is negligible

this strong force on a very small particle, the proton

would have been fly out from the nucleus

But there is another stronger force, called strong nuclear

•

force (F n) which is more strong than electrostatic force

that acts and holds nucleons (p – p, p – n, n – n) closely.

nuclear force

It is the most strong force in the universe and it acts

only between the nucleons

The properties of nuclear force are as follows:

Very short range :

or 4 fermi More than this distance, nuclear force is

almost zero

Very much depends upon distance :

in distance may cause large change in nuclear force

while electrostatic force remains almost unaffected

Independent of charge :

well as between p – p and also between n – p.

Spin dependent :

having same sense of spin than between p nucleons

having opposite sense of spin

The radius of a nucleus depends only on its mass

number A according to the relation r = r0A1/3, where r0

is a constant having a value of 1.2 fermi

isotopes, isobars and isotonesIsotopes : Isotopes of an element are nuclides having

same atomic number Z but different mass number A (or different neutron number N) Isotopes of an element

have identical electronic configuration and hence, identical chemical properties

11

12 13 116 126 146

H, H, H and C, C, C etc are isotopes

Isobars : Nuclides having same mass number A but

different atomic number Z are called isobars.Isobars

have different chemical properties In isobars, number

of protons Z as well as neutrons N are different but total nucleon (or mass) number A = N + Z is the same

2 3

2 3 6 14 7 14

He and H, C, N are isobars

Isotones : Nuclides with different atomic number Z and

different mass number A but same neutron number are called isotones Thus, for isotones N = (A – Z) is same

1 3 2 4

80 198 79 197

H, He and Hg, Au are examples of isotones

mass defect and Binding energyMass defect : The difference in mass of a nucleus and its

constituents is called the mass defect of that nucleus

Thus, mass defect, DM = Zm p + (A – Z)m n – M where M is the mass of given nucleus.

Packing fraction : Packing fraction of an atom is the

difference between mass of nucleus and its mass number divided by the mass number

A

Binding energy : The energy equivalent of the mass defect

of a nucleus is called its binding energy

Thus, binding energy

DE b = DMc2 = [Zm p + (A – Z) m n – M]c2

If masses are expressed in atomic mass units, then

= [Zm p + (A – Z)m n – M] × 931.5 MeV Binding energy per nucleon (DE bn) is the average energy needed to separate a nucleus into its individual nucleons Then

A

bn= b

Binding energy curve

It is the curve drawn between binding energy per nucleon and mass number as shown in the figure

The main features of the curve as follows :

2 4 6 8 10

constant, i.e practically independent of the

atomic number for nuclei of middle mass number

(30 < A < 170) The curve has a maximum of about

8.75 MeV for A = 56 and has a value of 7.6 MeV

of a body is measure of its energy content

Nuclei with high binding energies are very

•

stable as it takes a lot of energy to split them

Nuclei with lower binding energies are easier to

•

split

radioactivity

Spontaneous emission of radiations from the

nucleus is known as radioactivity and substances

showing this property are called radioactive

substances

Only unstable nuclei exhibit this property A particular

nuclide (element) can emit only a particular type of

radiations at a time, according to its requirement of

stability

Radioactivity was discovered by Becquerel therefore the

radiation also called Becquerel radiations

Later on Marie Curie and Pierre Curie discovered many

other radioactive substances

nature of radioactive radiationsrutherford’s experiment

Rutherford put a sample of radioctive substance

•

in a lead box and allow the emission of radiations through a small hole only When the radiation enter into the external electric field, they split into three parts

Radiations which deflect towards negative

•

plate are called a-rays as shown in figure These rays are stream of positive charged particles

Radiations which deflect towards positive plate are

•

called b-rays as shown in figure These are stream

of negative charged particles

Radiations which remain undeflected are called

•

g-rays as shown in figure These are electromagnetic wave or photons (electrically neutral) alike light rays

Physics for you |FEBRUARY ‘16 41

Comparison of a, b and g rays

Relative ionizing

power Very high, nearly 100 times of a-rays Small, nearly 100 times of g-rays Very small

Effect on ZnS

Nature of product Product obtained by the loss

of one a-particle has atomic number less by 2 units and mass number by 4 units

Product obtained by the loss

of one b-particle has atomic number more by 1 unit, without any change in mass number

There is no change in atomic number as well as mass number

law of radioactive decay

dN

dt = −lN t( ) or N t N e( )= 0 −lt

where l is the decay constant and N(t) is the number of

radioactive nuclei present at time t.

Half-life of a radioactive substance is given by

Activity : The number of disintegrations occurring

in a radioactive substance per second and it is given

where R0 = lN0 is the decay rate at t = 0 and R = Nl.

alpha, Beta and Gamma decayalpha decay

A nucleus that decays spontaneously by emitting an alpha particle (a helium nucleus 24 He) is said to undergo alpha decay The alpha decay is represented by

Z

A

Z A

X→ −−24Y + 24He +Q

where Z A X is the parent nucleus and Z A−−24Y is the daughter

nucleus and Q is the energy released in the decay.

In a alpha decay disintegration energy Q is given by

In beta minus decay (b–), a neutron is transformed into

a proton and an electron and antineutrino is emitted

In beta minus decay, disintegration energy Q is given by

Q = [m X – m Y ]c2

In beta plus decay (b+), a proton is transformed into

neutron and positron and neutrino is emitted

p → n + e+ + ν

where e+ is the positron and ν is the neutrino

The beta plus decay is represented by

the beta decay vary continuously from zero to a

kinetic energy K.E.max of an electron or positron

must equal to the disintegration energy Q.

substance can have unique and discrete values

Physics for you |FEBRUARY ‘16 43

SELFCHECK

10 Let Nb be the number of b particles emitted by

1 gram of Na24 radioactive nuclei (half life = 15 hrs)

where A is the target nucleus, a is the impinging particle, B

and b the products, Q is the energy released in the process.

The nuclear reaction is represented by notation A(a, b)B.

Q value of nuclear reaction,

Q = (m A + m a – m B – m b )c2

If Q is positive, the reaction is exothermic and if Q is

negative, the reaction is endothermic.

nuclear fission

It is the phenomenon of splitting a heavy nucleus into

two or more smaller nuclei

The nuclear fission of 92U235 is represented as

92U235 + 0n1 → 56Ba141 + 36Kr92 + 3 0n1 + Q

The value of the Q is 200 MeV per fission reaction.

Nuclear chain reaction : Under suitable conditions, the

three secondary neutrons may cause further fission of

U235 nuclei and start what is known as nuclear chain

reaction The nuclear chain reaction is controlled by

Neutron reproduction factor (K)

=rate of production of neutronsrate of loss of neutronsUncontrolled nuclear chain reaction is the basis of an atom bomb Controlled nuclear chain reaction is the basis of a nuclear reactor

Nuclear reactor : It is based on the phenomenon of

controlled nuclear chain reaction Moderators like heavy water, graphite, paraffin and deuterium slow down neutrons Rods of cadmium or boron serve as control rods Ordinary water and heavy water serve as coolants

nuclear fusion

It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus The nuclear fusion reaction of two deutrons is represented as

1H2 + 1H2 → 2He4 + 24 MeVTemperature ≈ 107 K are required for fusion to take place Nuclear fusion is a basis of hydrogen bomb

Stellar energy: It is the energy obtained from the sun

and stars The source of stellar energy is nuclear fusion

ansWer keys (self check)

Gamma Knife radiosurgery is becoming a very promising medical procedure for treating

certain problems of the brain, including benign and cancerous tumors, as well as blood

vessel malformations The procedure, which involves no knife at all, uses powerful, highly

focused beams of g-rays aimed at the tumor or malformation They (g-rays) are emitted by a

radioactive cobalt-60 source As figure illustrates, the patient wears a protective metal helmet

that is perforated with many small holes The holes focus the g-rays to a single tiny target within

the brain The target tissue thus receives a very intense dose of radiation and is destroyed, while

the surrounding healthy tissue is undamaged Gamma Knife surgery is non-invasive, painless,

and bloodless procedure that is often performed under local anesthesia Hospital stays are

70-90% shorter than with conventional surgery, and patients often return to work within a

few days

Gammarays

Target

Helmet

1 An a-particle moves in a circular path of radius

0.83 cm in the presence of a magnetic field of

0.25 Wb m–2 The de Broglie wavelength associated

with the particle will be

2 Photoelectric effect experiments are performed

using three different metal plates p, q and r having

work functions fp = 2.0 eV, fq = 2.5 eV and

fr = 3.0 eV, respectively A light beam containing

wavelengths of 550 nm, 450 nm and 350 nm with

equal intensities illuminates each of the plates The

correct I-V graph for the experiment is

(a)

q r V

I

r qp

V

3 A photon collides with a stationary hydrogen atom

in ground state in elastically Energy of the colliding

photon is 10.2 eV After a time interval of the order

of microsecond, another photon collides with same

hydrogen atom in elastically with an energy of

15 eV What will be observed by the detector?

(a) One photon of energy 10.2 eV and an electron

of energy 1.4 eV

(b) Two photons of energy 1.4 eV

(c) Two photons of energy 10.2 eV

(d) One photon of energy 10.2 eV and another

photon of 1.4 eV

4 A radioactive material decays by simultaneous

emission of two particles with respective half-lives

1620 and 810 years The time, (in years), after which

one-fourth of the material remains is

5 Rest mass energy of an electron is 0.7 MeV If the velocity of the electron is 0 51 ,c then kinetic

energy of the electron is nearly

6 Find the ratio of de Broglie wavelength of molecules

of hydrogen and helium which are at temperatures 27°C and 127°C, respectively

8 In a series of photoelectric emission experiments

on a certain metal surface, possible relationships between the following quantities were investigated: threshold frequency u0, frequency of incident

light u, light intensity P, photocurrent I, maximum kinetic energy of photoelectrons Tmax Two of these

quantities, when plotted as a graph of y against x,

give a straight line through the origin

Which of the following correctly identifies x and y

with the photoelectric quantities?

laser source, used in the school laboratory, shines from a spacecraft of mass 1000 kg Estimate the time needed for the spacecraft to reach a velocity of 1.0 km s–1 from rest

Physics for you |february ‘16 45

10 A hydrogen atom in the ground state absorbs

10.2 eV of energy The orbital angular momentum

of the electron is increased by

l1 and l2 are the de-Broglie wavelengths of the

particle, when 0 ≤ x ≤ 1 and x > 1 respectively If the

total energy of particle is 2E0, the ratio l

l12 will be

12 In the diagram a graph between the intensity of X-rays

emitted by a molybdenum target and the wavelength

is shown, when electrons of 30 keV are incident on

the target In the graph one peak is of Ka line and the

other peak is of Kb line

(a) First peak is of Ka line at 0.6 Å

(b) Highest peak is of Ka line at 0.7 Å

(c) If the energy of incident particles is increased,

then the peaks will shift towards left

(d) If the energy of incident particles is increased,

then the peaks will shift towards right

13 Two radioactive nuclei P and Q, in a given sample

decay into a stable nucleus R At time t = 0, number

of P species are 4 N0 and that of Q are N0 Half-life

of P (for conversion to R) is 1 minute whereas that of

Q is 2 minutes Initially there are no nuclei of

R present in the sample When number of nuclei of

P and Q are equal, the number of nuclei of R present

in the sample would be

14 The sun radiates energy in all directions The average

radiations received on the earth surface from the

sun is 1.4 kW m–2 The average earth sun distance is 1.5 × 1011 metres The mass lost by the sun per day is(a) 4.4 × 109 kg (b) 7.6 × 1014 kg

(c) 3.8 × 1012 kg (d) 3.8 × 1014 kg

15 A nucleus with mass number 220 initially at rest

emits an a-particle If the Q value of the reaction is 5.5

MeV, the kinetic energy of the a-particle will be

16 A proton, a neutron, an electron and an a-particle have same energy Then their de-Broglie wavelengths compare as

(a) lp = ln > le > la (b) la < lp = ln < le(c) le = lp > ln > la (d) le = lp > la > ln

on a surface with a spacing d = 0.1 nm The first

maximum of intensity in the reflected beam occurs

at q = 30° The kinetic energy E of the beam is

18 An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E

The rate of change of de-Broglie wavelength of the

electron at time t (ignoring relativistic effect) is

to the cube of principal quantum number

IV Binding force with which the electron is bound

to the nucleus increases as it shifts to outer orbits

(c) I, II and III (d) II, III and IV

20 A beam of 13.0 eV electrons is used to bombard gaseous hydrogen The series obtained in emission spectra is/are

(c) Paschen series (d) All of these