SAT II Physics (Gary Graff) Episode 2 Part 1 potx

A white spot from the light beam ap-peared on the wall, and when Newton placed a prism into the path ofthe light beam, the white light disappeared and was replaced by what seven-is calle

CHAPTER 6

MODERN PHYSICS

PARTICULATE THEORY OF LIGHT

Sir Isaac Newton studied the continuous spectrum early in the teenth century He passed a beam of sunlight through narrow open-ings into a darkened room A white spot from the light beam ap-peared on the wall, and when Newton placed a prism into the path ofthe light beam, the white light disappeared and was replaced by what

seven-is called a continuous spectrum Newton noticed that the spectrum

was displaced slightly to the side of the light path He also observedthat the colors of the spectrum always appeared in a continuous band

in the same order The red light always appeared closest to theoriginal path of the light path, followed by orange, yellow, green, blue,and violet, which was always deflected most from the original path ofthe white light Newton recognized the bending of light (refraction)

as the same process that occurred with water waves Since the lightexhibited the same characteristics as water waves, the wave nature oflight was easy to visualize

Max Planck spent the years from about 1890 to 1905 reviewingthe results of Heinrich Hertz’s experiments regarding the radiation ofhot objects Planck noticed that the results of Hertz’s experimentscould not be explained in terms of wave theory, but could be ex-plained if the energy in radiation was carried in bundles or packets of

light, which he called quanta Planck theorized that the energy of the

light was proportionally related to the frequency, which meant thatthe higher the frequency of the light, the higher the energy of thelight Planck related his idea to the equation:

E = hf where E is the energy, f is the frequency, and h is Planck’s constant.

It’s value is 6.6 × 10–4 J•S.

Planck’s theory is useful because it relates the frequency of light

to the energy carried by the light The light quanta suggested byPlanck also shows that the line spectra emitted by energized atomsare a unique set of frequencies that can best be explained by theparticle theory (quanta) of light

Albert Einstein used the quantum theory expressed by Planck toexplain the photoelectric effect Through the photoelectric effect,electrons are energized by light that is shined onto the surface of aphotosensitive metal Einstein concluded that when an electron isstruck by a quanta of light (he called them photons) the electrongains enough energy to be ejected from the surface of the metal.Einstein reached this conclusion through an experiment where anegatively charged zinc plate emitted photoelectrons when struck by

an ultraviolet light, but not when visible light was used Further, whenthe plate was given a positive charge neither ultraviolet light norvisible light produced electron emission

Only light of the correct frequency could energize electrons fromthe surface of the metal Einstein stated that the electron absorbs orreleases energy one photon at a time He concluded that the higherthe frequency of the light, the greater its energy Thus, yellow lightcarries less energy than green or blue light In fact, the continuousspectrum is arranged in order by the energy of the frequency of thelight Einstein also showed that light possessing the largest amount ofenergy (the highest frequencies) is refracted the most by a prism.Thus, the more a beam of light is refracted by a prism, the moreenergy it possesses

We can compare the energy content of different photons byusing Planck’s equation to calculate the energy of photons of thedifferent colors The relationship between different light colors andtheir wavelengths is given in the chart below We’ll use it to help usperform the energy calculations for the different light colors

Now let’s calculate the energy of a photon of yellow light

E hf

J s Joules

PHOTOELECTRIC EFFECT

When we apply the quantum theory to the photoelectric effect, wemust realize that the KE of the ejected electron is directly related to

the energy it receives from the incident photon, minus the energy (W)

required to remove the electron from the surface of the metal (workfunction)

KE = hf –W 1/2 mv2 = hf–W

We see from the equation above that no electrons can be emittedfrom the surface of the metal unless the product of the frequency and

Planck’s constant is greater than the work function (W) In addition,

the kinetic energy of the ejected electron depends upon the frequency

of the photon and the work function of the metal As we have seenfrom the chart of the color, frequency, and photon energy, the energy

of the emitted photon is very small A more convenient method tomeasure the energy of the electron (whether it absorbs or emits thephoton) is to use an energy scale that is of the same magnitude as the

electron: the electron volt (eV) The electron volt is defined as the

quantity of work required to move an electron through a potentialdifference of 1 volt (Remember, 1 volt is equal to 1 Joule percoulomb of charge.)

Using this relationship, we will go back and recalculate the

energy of a photon of yellow light in eV.

The following problem illustrates the usefulness of the electronvolt when working with photo-voltaic materials.

Example

The work function of a metal is given as 2.46eV What is the kinetic

energy of the photons ejected from the surface of the metal when alight with a frequency of 8.2 × 1014 Hz shines on the metal?

Solution

The work function is the minimum energy required to dislodge anelectron from the surface of the metal, and it must be subtracted fromthe energy of the incident photons

The work function of a metal is equal to the energy of thephoton that energizes the electron

E = hf

The threshold frequency of the light can be calculated if the workfunction is known

J s eV

λλλλ

m/s

m/wavve

Again we obtain a value that places the light into the ultravioletrange

We’ll do one more problem This time, try to find the workfunction of a light that approaches the infrared range The limit ofvisible light in the red range is 3.9 × 1014 Hz Calculate the work

E = hf

Bohr thought the electrons in their orbits would emit energybased upon their distance from the nucleus He stated the relation-ship as:

block of carbon, Compton noticed that he kept obtaining a smallchange in wavelength, which he decided to investigate He consideredthe wavelength to be approximately the size of a single atom, andsince the X-ray energy is so large, the energy needed to knock a singleatom of the carbon is negligible compared to the total energy of theX-ray The X-ray photon was deflected in the collision with the carbonelectron, while the electron had a velocity impressed on it Applyingthe laws of conservation of momentum and energy to the collisionsproduced the conclusion that the X-ray photon does have momentum.Louis de Broglie investigated one of Bohr’s theories while con-templating the dual nature of the electron He found that by assum-ing the electron to be capable of having wave properties, he couldexplain one of Bohr’s assumptions about electrons de Broglie postu-lated that moving particles could have wave properties, leading to thefollowing:

deBroglie wavelength = λ = h

mv

where mv = momentum of a photon

Should we attempt to calculate the wavelength of a large object(not on atomic scale), we find the wavelength of the object to be verysmall, which makes the waves undetectable

Example

A 900 kg automobile is being driven down the road with a velocity of

30 m/s What de Broglie wavelength would the car emit?

Solution

λλλ

J s Kg

REFERENCE FRAMES

One of the most difficult concepts to accept is that the length of ameter stick, or the mass of an apple, or even the ticking of a clock canchange without any of the items being broken, damaged, or in disre-pair The relative length, mass, or rate of movement of these items isexpected to remain the same relative to other like items The key

word is relative You see, the comparison only remains the same for

the objects when they are in the same reference frame

What is a reference frame? Just think of driving down the road at

a velocity of 70 km/hr When you pass a telephone pole, the phone pole flashes by in an instant Your reference frame is the inside

tele-of the car The windows, the dash, the seats, and yourself are notmoving in reference to one another Outside, the rest of the worldflies by at 70 km/hr

Now let’s suppose you are seated in the backseat of the movingvehicle with all the windows closed You toss a wad of paper fromyour side of the car to the passenger on the other side of the car Thatperson catches the paper wad and throws it right back to you As far

as you’re concerned, the paper wad flew straight across the back ofthe car in both instances

Let’s suppose the top of your car is transparent, and an observer

in a tall tower is looking down on the events as they take place in thecar To that person, your game of catch looks like this:

You and the person in the tower see the events in the back of thecar in a very different way For purposes of clarity, the numbers weuse in the following discussion will be exaggerated a little, but theidea will still be relevant and may be easier to grasp Your speed inthe car (70 km/hr) is about 20 m/s Let’s say you and your friend bothtoss the paper wads with a speed of 10 m/s The car (remember theexaggeration) is 10 m wide From your perspective, the paper wadtakes 1 second to cross the width of the car Everything inside the car

is just fine as far as you’re concerned The car is 10 m wide, youthrew the paper at 10 m/s, and it takes 1 second for the paper tocross the interior of the car

The observer outside the car sees things differently

During the 1 second time span, the paper wad flew from one

(20m)(1 sec) = 20m The width of the car is 10m, so the path of thepaper wad as seen by the outside observer is the hypotenuse of thetriangle shown To the outside observer it is 22.4 m long, and thevelocity with which you throw the paper wad is considerably fasterthan 1 m/s In fact, the observer measures the velocity of the paperwad to be 22.4 m/s.

When discussing relativity, you must always be aware that the reference frame is the key. One of the conclusions we can drawabout reference frames is that the laws of mechanics are the same inall reference frames moving at constant velocity with respect to oneanother In addition, all motion is relative to some reference frame.Now that we have discussed reference frames, let’s proceed toEinstein’s theory Einstein postulated that

1 all the laws of physics are the same for all observers moving atconstant velocity with respect to one another

2 the speed of light in a vacuum is the same for all observersregardless of the motion of the source of light or the motion ofthe observer

SIMULTANEITY

Let’s first look at the second postulate of Einstein’s theory We willreturn to the car, but instead of a wad of paper, we will use a beam oflight, and instead of a second passenger we will use a mirror toreflect the light In addition, the velocity of the car will be increased

to 75 the speed of light Inside the car, however, nothing changes asfar as you are concerned Things are just as they were when youtraveled at 70 km/hr

Example

The light beam you send from your side of the car crosses the car,strikes the mirror, and returns to your side of the car The light travelsthe ten meters to the mirror and the 10 meters from the mirror in6.67 × 10–8 s at a velocity of 3 × 108 m/s

Solution

The question is, what does the outside observer see? The answer? Theoutside observer sees the light beam complete the trip in the sameamount of time as you do This means that the light traveled either alonger distance in the same period of time, thereby breaking one ofthe postulates of the theory of relativity, or, unlikely as it may seem,the outside observer was in a frame of reference where time passedmore quickly

The reason for the conclusion is that the person in the car seesthe light travel from the origin, strike the mirror, and return to him,which is expressed as:

=2 for the round trip

The outside observer sees things differently

The path length the outside observer sees is considerably longerthan the one seen by the person in the car This means:

Since c must be the same for both parties, ∆t and ∆t0 cannot bethe same

Einstein related the difference in time for the two observers withthe relativistic equation:

0 the time on the stationary clock

Let’s take a look at what this means in terms of the two people.Suppose the person in the car is moving at 75 the speed of light.What time will pass on the rider’s clock if the observer in the towermeasures a 30-minute time span?

The rider in the car moving at 75 c only measures a time of 19.8

minutes on a clock, while the outside observer measures a time of 30minutes

The relativistic equation shows that the clock in the moving carmoves more slowly This fact leads to the following statement abouttime dilation

Clocks in a moving reference frame run more slowly

Now that we have seen the time dilation part of Einstein’s theory

of relativity, the question arises about mass and other quantities.According to Einstein’s relativistic equation, an object that has alength of 1 meter will be shorter to an outside observer Let’s take alook at length

LENGTH CONTRACTION

Example

A hypothetical traveler going to Alpha Centauri will be in a spaceshipthat can travel at 85 the speed of light How will the traveler experi-ence the trip compared to an observer on earth?

Solution

We already know the clocks on the spaceship will move more slowly.That means the time on the earth ticks away faster Substituting thedistances into the relativistic factor instead of time yields the answer

d t = distance according to the spaceship traveler

d0 = distance traveled to and from the star according to the knownmeasurement

The distance to Alpha Centauri is approximately 4.5 light years,

so we multiply by 2 to find the round trip distance

Substituting and solving, we have:

c d

The traveler finds the distance to the star and back again to be4.74 light years instead of the 9 light years measured by earthboundobservers.

RELATIVISTIC MASS—ENERGY RELATION

As stated earlier, the postulates of Einstein’s theory of relativity tell usthat no object can be accelerated beyond the speed of light Accord-

ing to Newton’s laws of motion, any object could be accelerated to

the speed of light if given enough time To be consistent with thelaws of momentum, Einstein determined that the mass of an objectmust increase when its velocity increases This led him to conclude:

v c

1

This means an electron moving at nearly the speed of light gains

mass Let’s use 98c, for example,

.( )

KE = (m–m0)(c2)Notice the change in mass requires more and more energy toincrease the speed of the object This finally leads to the equation:

E = mc2

This is perhaps Einstein’s most famous equation, and it is theequation that eventually lead to the development of nuclear weaponsand nuclear power