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948 41 Quantum Mechanics CHAPTER OUTLINE 41.1 The Wave Function 41.2 Analysis Model: Quantum Particle Under Boundary Conditions 41.3 The Schrödinger Equation 41.4 A Particle in a Well

948

41

Quantum Mechanics CHAPTER OUTLINE

41.1 The Wave Function

41.2 Analysis Model: Quantum Particle Under Boundary Conditions

41.3 The Schrödinger Equation

41.4 A Particle in a Well of Finite Height

41.5 Tunneling Through a Potential Energy Barrier

41.6 Applications of Tunneling

41.7 The Simple Harmonic Oscillator

* An asterisk indicates a question or problem new to this edition

ANSWERS TO OBJECTIVE QUESTIONS

OQ41.1 Answer (b) Fewer particles are reflected as the height of the

potential barrier decreases and approaches the energy of the particles By Equations 41.22 and 41.23, the transmission coefficient

T ≈ e −2CL, where C = 2m U − E( )  , increases as U − E decreases, so the reflection coefficient R = 1 − T ≈ 1 − e −2CL decreases as U − E

decreases

OQ41.2 The ranking is answer (b) > (a) > (c) > (e) > (d) From Equation 41.14,

consider the quantity

OQ41.3 (a) True Examples: An electron has mass and charge, but it can

also display interference effects

(b) False An electron has rest energy E R = m e c2 (c) True A moving electron possesses kinetic energy

(d) True p = m e u

(e) True

OQ41.4 (a) True Examples: A photon behaves as a particle in the

photoelectric effect and as a wave in double-slit interference (b) True A photon cannot have rest energy (mass) because it is never at rest: it travels at the speed of light

(c) True E = hf

(d) True p = E/c

(e) True

OQ41.5 Answer (d) The probability of finding the particle is at the antinodes

(places of greatest amplitude) of the standing wave

OQ41.6 Compare the ground state wave functions in Figures 41.4 and 41.7 in

the text In the square well with infinitely high walls, the particle’s

simplest wave function has strict nodes separated by the length L of the well The particle’s wavelength is 2L, its momentum h/2L, and its energy p2/2m = h2/8mL2 In the well with walls of only finite height, the wave function has nonzero amplitude at the walls, and it extends outside the walls

(i) Answer (a) The ground state wave function extends somewhat outside the walls of the finite well, so the particle’s wavelength

is longer

(ii) Answer (b) The particle’s momentum in its ground state is

smaller because p = h/λ and the wave function has a larger

wavelength

(iii) Answer (b) The particle has less energy because is has smaller

momentum

OQ41.7 Answer (e) From the relation between the square of the wave

function and the probability P of finding the particle in the interval

Δx = (7 nm − 4 nm) = 3 nm, we have

OQ41.8 Answer (a) Because of the exponential tailing of the wave function

within the barrier, the tunneling current is more sensitive to the width of the barrier than to its height Notice that the exponent term

CL in the transmission coefficient T ≈ e −2CL, where

C = 2m U − E( )  , decreases more if L decreases than if U decreases

by the same percentage

OQ41.9 Answer (c) Other points see a wider potential-energy barrier and

carry much less tunneling current

OQ41.10 Answer (d) The probability of finding the particle is greatest at the

place of greatest amplitude of the wave function The next most

likely place is point b, after that, points a and e appear to be equally probable The particle would never be found at point c

ANSWERS TO CONCEPTUAL QUESTIONS

CQ41.1 Consider the Heisenberg uncertainty principle It implies that

electrons initially moving at the same speed and accelerated by an

electric field through the same distance need not all have the same

measured speed after being accelerated Perhaps the philosopher could have said “it is necessary for the very existence of science that the same conditions always produce the same results within the uncertainty of the measurements.”

CQ41.2 Consider a particle bound to a restricted region of space If its

minimum energy were zero, then the particle could have zero momentum and zero uncertainty in its momentum At the same time, the uncertainty in its position would not be infinite, but equal to the

width of the region In such a case, the uncertainty product ΔxΔp x

would be zero, violating the uncertainty principle This contradiction proves that the minimum energy of the particle is not zero

CQ41.3 The motion of the quantum particle does not consist of moving

through successive points The particle has no definite position It can sometimes be found on one side of a node and sometimes on the other side, but never at the node itself There is no contradiction here, for the quantum particle is moving as a wave It is not a classical particle In particular, the particle does not speed up to infinite speed

to cross the node

CQ41.4 (a) ψ (x) becomes infinite as x → ∞

(b) ψ (x) is discontinuous and becomes infinite at x = π/2, 3π/2,…

CQ41.5 A particle’s wave function represents its state, containing all the

information there is about its location and motion The squared absolute value of its wave function tells where we would classically think of the particle as spending most its time Ψ2 is the probability distribution function for the position of the particle

CQ41.6 In quantum mechanics, particles are treated as wave functions, not

classical particles In classical mechanics, the kinetic energy is never negative That implies that E ≥ U. Treating the particle as a wave, the Schrödinger equation predicts that there is a nonzero probability that

a particle can tunnel through a barrier—a region in which E < U

CQ41.7 Both (d) and (e) are not physically significant Wave function (d) is

not acceptable because ψ is not single-valued Wave function (e) is not acceptable because ψ is discontinuous (as is its slope)

CQ41.8 Newton’s 1st and 2nd laws are used to determine the motion of a

particle of large mass The Schrödinger equation is not used to determine the motion of a particle of small mass; rather, it is used to determine the state of the wave function of a particle of small mass

In particular, the states of atomic electrons are confined-wave states whose wave functions are solutions to the Schrödinger equation Anything that we can know about a particle comes from its wave function

SOLUTIONS TO END-OF-CHAPTER PROBLEMS

P41.1 (a) The wave function,

Section 41.2 Analysis Model: Quantum Particle

Under Boundary Conditions

P41.5 (a) The energy of a quantum particle confined to a line segment is

E n = h2n2

8mL2 Here we have for the ground state

E2 = 4E1= 2.05 MeV and E3 = 9E1= 4.62 MeV(b) They do; the MeV is the natural unit for energy radiated by an atomic nucleus

Stated differently: Scattering experiments show that an atomic nucleus is a three-dimensional object always less than 15 fm in diameter This one-dimensional box 20 fm long is a good model in energy terms

P41.6 From Equation 41.14, the allowed energy levels of a particle in a box is

Substituting numerical values,

( )2 − 3( )2 = 4.71 nm The wavelengths produced by all possible transitions are:

Transition 4 → 3 4 → 2 4 → 1 3 → 2 3 → 1 2 → 1

λ (nm) 4.71 2.75 2.20 6.59 4.12 11.0

P41.8 The energy of the photon is

E= hc

λ =

1 240 eV⋅ nm6.06 mm

1 mm

106nm

⎛

⎝⎜ ⎞⎠⎟ = 2.05 × 10−4eVThe allowed energies of the proton in the box are

The smallest possible energy for a transition between states is from

n = 1 to n = 2, which has energy

ΔE n = 2.05 × 10( −4eV) (22− 12) = 6.14 × 10−4eV The photon does not have enough energy to cause this transition The

photon energy would be sufficient to cause a transition from n = 0 to

n = 1, but the n = 0 state does not exist for the particle in a box

P41.11 From Equation 41.14, the allowed energy levels of a particle in a box is

E = ΔE n = E2− E1= 2( )2E1− E1= 3E1= 6.14 MeV(b) The wavelength of the photon is

λ= hc

E = 1 240 eV⋅ nm6.14× 106 eV

= 2.02 × 10−4 nm= 2.02 × 10−13 m= 202 × 10−15 m= 202 fm(c) This is a gamma ray, according to the electromagnetic spectrum chart in Chapter 34

P41.12 The ground state energy of a particle (mass m) in a 1-dimensional box

P41.13 E1 = 2.00 eV = 3.20 × 10–19 J For the ground state,

*P41.15 (a) The energies of the confined electron are

E n = h2

8m e L2 n2 Its energy gain in the quantum jump from state 1 to state 4 is

(b) Let λ represent the wavelength of the photon emitted: ′

(b) Its energy is all kinetic, so

= 3.99 × 10−2

(d)

In the n= 2 graph in the text’s Figure 41.4(b), it is more probable

to find the particle either near x = L 4 or x = 3L 4 than at the

center, where the probability density is zero Nevertheless, thesymmetry of the distribution means that the average position is

locations are at the antinodes of the standing wave pattern n = 3,

which has three antinodes that are equally spaced, one at the

center, and two a distance L/4 from either end

P41.21 (a) The probability of finding the electron between x = 0 and

2π3

⎛

⎝⎜ ⎞⎠⎟ = 13−0.8662π = 0.196(b) Classically, the particle moves back and forth steadily, spending equal time intervals in each third of the line The classical

probability is 0.333, which is significantly larger

3− 0 = 0.333

The probability is 0.333 for both classical and quantum models

P41.22 (a) From Equation 41.13,

The wave function is zero for x < 0 and for x > L The

probability at  = 0 must be zero because the particle

is never found at x < 0 or exactly at x = 0 The probability

at  = L must be 1 for normalization: the particle is always

found somewhere in the range 0< x < L.

(d) The probability of finding the particle between x = 0 and x =  is 2

3, and between x =  and x = L is

This equation for u can be solved by homing in on the solution

with a calculator, the result being

2π3

2π3

⎛

⎝⎜ ⎞⎠⎟

= 1

3− 34π

x= 2L

3 and x = L is the same, 0.196 Therefore, the probability of finding it in the range

Section 41.3 The Schrödinger Equation

The wave function ψ = Ae i kx( −ωt) is a solution to the Schrödinger

equation if equation [3] is true Both sides depend on A, x, and t in the

same way, so we can cancel several factors, and determine that we have a solution if

where K is the kinetic energy Therefore, the given wave function

does satisfy Equation 41.15

P41.25 (a) Given the function

ψ x( )= Acoskx + Bsin kx Its derivative with respect to x is

∂ψ

∂x = −kAsin kx + kBcos kx

And its second derivative is

∂2ψ

∂x2 = −k2A cos kx − k2B sin kx

= −k2(A cos kx + Bsin kx)= −k2ψThe Schrödinger equation is satisfied if

and the probability density is

Schrödinger equation to isolate the potential energy function gives

(b) U(x) is sketched in ANS FIG P41.27(b)

(b) Note that the wave function ψ x( ) is an even function; therefore,

we may write the normalization condition as

P41.29 (a) For n = 4, the wave function has two maxima and two minima

(four extrema), as shown in the left-hand panel of ANS FIG P41.29

(b) For n = 4, the probability function has four maxima as shown in

the right-hand panel of ANS FIG P41.29

ANS FIG P41.29 P41.30 (a) See ANS FIG P41.30(a)

ANS FIG P41.30(a)

(b) The wavelength inside the box is 2L The wave function

penetrates the wall, but the wavelength of the transmitted wave traveling to the left is the same, 2L , because U = 0 on both sides

of the wall, so the energy and momentum and, therefore, the wavelength, are the same

P41.31 The decay constant for the wave function inside the barrier is:

P41.33 The original tunneling probability is T = e −2CL, where

P41.34 With the wave function proportional to e –CL, the transmission

coefficient and the tunneling current are proportional to ψ 2, to e –2CL Then,

Section 41.7 The Simple Harmonic Oscillator

P41.36 (a) The wave function is given by ψ = Axe −bx2

m ψ −2b22

2ψ = −1

2mω2x2ψ + Eψ

For this to be true as an identity, the coefficients of like terms

must be the same for all values of x So we must have both

P41.37 The longest wavelength corresponds to minimum photon energy,

which must be equal to the spacing between energy levels of the

oscillator From E = ω , we have

P41.38 The longest wavelength corresponds to minimum photon energy,

which must be equal to the spacing between energy levels of the oscillator, which is (from Equation 41.28)

Substitute for u1 and u2 :

of the equivalent particle is a negative constant times the displacement from equilibrium

E = p x22m + k

2x

2 = p x22m +k

2m+ k2

8 ( )−1 1

Δp x

( )4 = 0Then

k

m+ 4

k

m = 2

k m

p= h

λ =

6.626× 10−34 J⋅ s2.00× 10−10 m = 3.31 × 10−24 kg⋅ m/s (c) And its energy is

P41.46 If we had n = 0 for a quantum particle in a box, its momentum would

be zero The uncertainty in its momentum would be zero The uncertainty in its position would not be infinite, but just equal to the width of the box Then the uncertainty product would be zero, to violate the uncertainty principle The contradiction shows that the quantum number cannot be zero In its ground state the particle has some nonzero zero-point energy

P41.47 T = e –2CL, where

C= 2m U( − E)

 and where m is in kilograms, and U

and E are in joules

P41.48 From Equation 41.14, the energy levels of an electron in an infinitely

deep potential well are proportional to n2 If the energy of the ground

state, n = 1, is E1 = 0.300 eV, the energy levels of the states n = 2, 3, and

(a) For the transition from the n = 3 level to the n = 1 level, the

electron loses energy

In like manner, we find for 3 to 2: ΔE = 1.50 eV, and λ = 827 nm, infrared

for 4 to 1: ΔE = 4.50 eV, and λ = 275 nm, ultraviolet

for 4 to 2: ΔE = 3.60 eV, and λ = 344 nm, near ultraviolet

for 4 to 3:

ΔE = 2.10 eV, and λ = 590 nm, yellow-orange visible

P41.49 (a) From E = hf, the frequency is

λ = c

f =3.00× 108 m/s4.35× 1014 Hz = 6.89 × 10−7 m= 689 nm

The uncertainty is 165 peV or more

P41.50 Suppose the marble has mass 20 g Suppose the wall of the box is 12

cm high and 2 mm thick While it is inside the wall,

U = mgy = 0.02 kg( ) (9.8 m/s2) (0.12 m)= 0.023 5 J and

E = K = 1

2mu

2 = 1

2(0.02 kg) (0.8 m/s)2 = 0.006 4 JThen,

C= 2m U( − E)

 = 2 0.02 kg( ) (0.017 1 J)

1.055× 10−34 J⋅ s = 2.5 × 1032 m−1

and the transmission coefficient is

the integral becomes (from integral tables)

2L are still valid

From the relativistic energy of the particle,

E1= h2

8mL2 = (6.626× 10−34 J⋅s)2

8 9.11× 10( −31 kg) (1.00× 10−12 m)2 = 6.02 × 10−14 J

Comparing this to K1, we see that this value is too large by 28.6%

P41.54 Looking at Figure 41.7, we see that wavelengths for a particle in a finite

well are longer than those for a particle in an infinite well Therefore, the energies of the allowed states should be lower for a finite well than for an infinite well As a result, the photons from the source have too much energy to be absorbed or, equivalently, the photons have a frequency that is too high In order to lower their apparent frequency

using the Doppler shift, the source would have to move away from the particle in the finite square well, not toward it

P41.55 (a) For a particle with wave function