Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
Problem Set 2: Photons
Solution
1. Consider the double slit set up that we considered in the class and recall its main
features. A coherent but faint source of light is shone on the double slit apparatus from
left. On Screen S, where light is finally detected, we have detectors available capabale
of measuring extremely small intensities of light. We observe that only flashes of light
are detected, always of same energy if we are using a single frequency light, which we
interpret as particles of light i.e., photons. When we collect the flashes, count and plot
them, we get the interference pattern. On the other hand, if one hole is closed, we get
a pattern that is just peaked in front of the open hole as shown in figure. If we open
the holes alternately, and wait for some time to collect the flashes, we just get the sum
of two such peaks and no interference. All this persists, even if the light is so faint
that we record flashes at a very slow rate such that there is only one photon crossing
the apparatus on average. Also, if we put detectors near the holes, these detectors
also only detect a flash of same size, only at one of the detctor at a time. In short, you
can not divide the size of the flash. Remember, the pattern on the screen is formed
only after collecting many flashes ( or particles of light i.e., photons).
Try to explain these features, using the four possibilities for a picture of the dual
particle and wave nature of light that were discussed in class. Explain which one
works and which one fails and why. If all of them fail, try to come up with a picture
of your own which could work.
Answer 1
Let us consider one of the four given possible pictures at a time and see if it explains
the results of double slit experiment.
Option A says that particles move in a straight line, just like classical bullets. Classical particles do not interfere, and would give the sum of two peaks (drawn in blue in
Figure 1) as the pattern on the screen when we leave both of the holes open. This is
against the observation we make (i.e. the pattern in red) and therefore, we know that
this picture is not correct for wave-particle duality.
Option B says that the wave particle duality can be understood by considering classi-
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
cal particles moving in wavy trajectories. Such particles would pass through the holes
and strike the screen as classical bullets do; there is no reason for them to interfere.
Hence, the interference pattern that we see, would not have been visible.
Option C says that photons can be visualized in the following way: consider an electromagnetic wave coming in from left; let us assume that it is divided into many
wavelets, then each of those wavelets is a photon. We know that as we reduce the
intensity of light to one photon at a time, each wavelet must pass through both holes
for the interference pattern to appear on the screen. But if we put a detector near
each of the holes, only one detector clicks (and it detects a flash of the same size as
the one we would detect at the screen). This problem cannot be resolved using this
picture of wave-particle duality, hence it is not true.
Option D says that associated to a photon is an electromagnetic wave such that the
photon and the wave move forward together. By using the same argument as the one
that we used to analyze option C, we know that as a single photon crosses the slits,
the wave must pass through both holes to give the interference pattern. But as we
detect the photon near the slits, we find that only one detector clicks. Hence, this
picture also does not explain the observations of Young’s double slit experiment.
As we will see in the course, the right way to think of wave-particle duality is through
Born’s interpretation. (In Born’s interpretation, to each particle is associated a wave,
whose intensity gives the probability of finding the particle at a particular position
and a particular time. But more on that, later!)
2. Consider a source of light that is capable of giving off light of single wavelength.
Suppose It is emitting light of wavelength λ1 = 1.40 × 10−7 m with a power of 0.005
Watts.
(a) How many photons are being emitted by the source in one second?
(b) Now suppose you double the wavelength but keep the power same. How many
photons are being emitted now?
(c) When the light with original wavelength λ1 is incident on a metal, some electrons
are emitted. The electrons in the metal need an unknown amount φ of energy,
at the very least, just to be liberated from the metal surface. Suppose it is found
that the highest kinetic energy of the emitted electrons is 4eV. Now suppose
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
you again double the wavelength and shine this light on the same metal surface.
What would be the maximum kinetic energy of the emitted electrons this time?
(d) Now you take the same material but halve the wavelength of the light being used.
What would be the maximum kinetic energy of the emitted electrons now?
Answer 2
We are given that,
Wavelength of light = λ1 = 1.40 × 10−7 m
Power = P1 = 0.005 Watts = 0.005 J/s.
(a) We want to calculate number of photons emitted by the source in one second.
For this purpose let’s at first calculate energy of one photon by using the relation,
E1 = hf1 ,
(1)
where h is the Planck’s constant with numerical value 6.63 × 10−34 J.s, and f1 is
the frequency of the emitted wave, can be calculated by using the equation,
c = f 1 λ1
c
.
⇒ f1 =
λ1
Therefore equation (1) can be written as,
hc
6.63 × 10−34 Js × 3 × 108 m/s
E1 =
=
λ1
1.40 × 10−7 m
−18
= 1.42 × 10
J.
From the given data we can see that power emitted is 0.005 Watt or in other words
energy emitted per second is 0.005 J, Therefore number of photons emitted per
second will be,
P1
0.005 J/s
=
E1
1.42 × 10−18 J
15
= 3.52 × 10 photons/s.
Number of photons = N1 =
Hence 3.52 × 1015 photons are being emitted per second.
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
(b) Now we are given that,
Wavelength of light = λ2 = 2λ1
Power = P2 = P1
Since number of photons emitted by the source are given by,
P
Pλ
=
E
hc
P2 λ 2
⇒ N2 =
hc
P1 × 2λ1
=
hc
= 2N1 .
N =
Hence doubling the wavelength would double the number of photons.
This makes sense since by increasing the wavelength, we are reducing the energy
being carried by each photon so to make up for the same power we would have
to squeeze more photons per second.
(c) Now we are given that,
Maximum kinetic energy of photon = K.E.Max.1 = 4 eV
Wavelength = λ2 = 2λ1 .
In photoelectric effect we read that maximum kinetic energy of photoelectrons
can be calculated by using the relation,
K.E.Max.1 = E1 − φ
⇒ φ = E1 − K.E.Max.1
From part (a) we know that E1 is the energy of photon striking the metal surface
and is given by, hc/λ1 , If wavelength is doubled the E2 = hc/λ2 = hc/2λ1 = E1 /2,
Thus maximum kinetic energy will become,
K.E.Max.2 = E2 − φ
E1
K.E.Max.2 =
−φ
2
E1
=
− (E1 − K.E.Max.1 )
2
E1
=
− E1 + K.E.Max.1
2
E1
= K.E.Max.1 −
·
2
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
From last expression we can see that the maximum kinetic energy is reduced by
an amount of E1 /2.
(d) By reducing the wavelength to half, energy will become 2E1 , therefore maximum
kinetic energy will be,
K.E.Max.3 = E3 − φ
K.E.Max.3 = 2E1 − φ
= 2E1 − (E1 − K.E.Max.1 )
= K.E.Max.1 + E1 .
Hence kinetic energy is increased by an amount of E1 .
3. Complete the derivation of recitation problem 3 to derive that the shift in wavelength
in Compton effect is given by
∆λ = λf − λi =
h
(1 − cosθ)
mc
where m0 is electrons rest mass, c is velocity of light and λf , λi are wavelengths of
scattered and incident X-rays respectively. If the light always strike the electrons like
billiard balls, then this shift should occur whenever light interacts with electrons (
radio signal transmission, light reflecting from a surface). Why did people not notice
this shift in wavelength of light before Compton and how come they beleive Maxwells
theory which predicts no shift in wavelength.
Answer 3
If the light strikes with an electron like billiard balls, you can use conservation of
momentum and conservation of energy to write equations (as you used to do in your
Mechanics course).
First, let us define all the variables that we will use in this calculation. Let pi and pf
be the momenta of incoming and scattered photons respectively, and let λi and λf be
their respective wavelengths. Then the energy of incoming and outgoing photons are
Ei = pi c = hc/λi and Ef = pf c = hc/λf respectively. Also, let the rest mass of the
electron be m0 and its momentum after the collision be denoted by p.
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
Before the collision, the total energy of the system is the sum of photon’s energy (pi c)
and the rest energy of the electron. After the collision, the total energy is the sum
of the photon’s energy (pf c) and the energy of the electron which we know from our
p
knowledge of relativity to be (pc)2 + (m0 c2 )2 . Hence the equation for conservation
of energy is:
pi c + m0 c2 = pf c +
p
(m0 c2 )2 + (pc)2
(2)
To write equations for conservation of momentum we assume that the motion of the
incoming photon is in the horizontal direction. We let θ and φ be the angles that
the scattered photon and the recoiled electron make with this direction. Then the
conservation of horizontal component of momentum implies that the following equation
is true.
pi = pcosφ + pf cosθ
(3)
Rearranging this equation gives:
pi − pf cosθ = pcosφ
(4)
Similarly, for the conservation of vertical component, we have the following equation:
0 = psinφ − pf sinθ
(5)
⇒ psinφ = pf sinθ
(6)
To eliminate φ we can square equations (4) and (6) and add:
p2i + p2f cos2 θ − 2pi pf cosθ + p2f sin2 θ = p2 cos2 φ + p2 sin2 φ
⇒ p2i + p2f − 2pi pf cosθ = p2
(7)
(8)
Now, to eliminate p2 we can plug in value of p2 from this equation into equation (2).
(pi − pf )c + m0 c2 =
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q
(m0 c2 )2 + p2i c2 + p2f c2 − 2pi pf c2 cosθ
(9)
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
Now, square on both sides to get
p2i c2 + p2f c2 − 2pi pf c2 + 2(pi − pf )m0 c3 = p2i c2 + p2f c2 − 2pi pf c2 cosθ
⇒ (pi − pf )m0 c3 = pi pf c2 (1 − cosθ)
pi − pf
1 − cosθ
⇒
=
pi p f
m0 c
h/λi − h/λf
1 − cosθ
⇒
=
2
h /(λi λf )
m0 c
h
⇒ λf − λi =
(1 − cosθ)
m0 c
(10)
(11)
(12)
(13)
(14)
which is the required expression for the shift in wavelength as a function of scattering
angle θ.
The reason that this shift was not observed earlier is because of the small value of
the factor h/m0 c that arises in the expression for Compton shift. By plugging in
h = 6.626 × 10−34 Js, m0 = 9.109 × 10−31 kg and c = 3 × 108 m/s we get that h/m0 c =
2.43 × 10−12 m. As you will see in Question 9, ∆λ is maximum when θ = π. Hence
the maximum value of ∆λ will be 4.86 × 10−12 m. Visible light has wavelength of
the order of 10−7 m. A change of the order of 10−12 m in the wavelength would go
undetected for the visible light. It was only at the dawn of 20th century that the
x-rays were discovered and their interaction properties with matter were studied that
the Compton shift in wavelength of X-rays was observed.
4. The energy gap between valance and conduction band in a material is 2.5eV what is
the minimum frequency of light that you must shine to make a solar cell made out of
this material to work?
Answer 4
We are given that,
Energy gap = EG = 2.5 eV.
We want to calculate the minimum frequency of the photon to make the material work.
As we know that the energy of the photon should be equal to the band gap energy,
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
therefore,
E = EG = hf
EG
2.5 eV
⇒f =
=
= 6.04 × 1014 s−1
−15
h
4.14 × 10
eVs
14
= 6.04 × 10 Hz
5. A light of wavelength λ = 750 nm is shone on a solar cell with a power of 1 mW
falling on a square centimeter of the cell. We see that a current of 50mA runs through
the circuit as a result of it. It is then found that if we double the power of light the
current also doubles, assuring us of a linear response of the current to the intensity of
light. Estimate how much current will run if we shine a light of λ = 400 nm on this
cell, again with a power of 1 mW falling on a square centimeter of the cell.
Answer 5 We are given that,
Wavelength of first light = λ1 = 750 nm = 750 × 10−9 m
Power of first lights = P1 = 1 mW = 10−3 W
Photoelectric current due to first light = I1 = 50 mA = 50 × 10−3 A
Wavelength of second light = λ2 = 400 nm = 400 × 10−9 m
Power of second lights = P2 = 1 mW = 10−3 W
Photoelectric current due to second light = I2 =?
As we know that the photoelectric current (I) is directly proportional to the number
of photons (N ) emitted, i.e.,
N ∝ I
⇒ N = KI,
where K is constant of proportionality. Therefore photoelectric current can be calculated by using the following relation,
I=
N
.
K
Let’s at first calculate N and K. In order to calculate K consider the first light, and
use the formula K = N/I. Since I is given for first light let’s calculate N We must
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
calculate energy of one photon at first, to find the number of photons emitted. As
energy of a photon is given by,
E1 = hf1
hc
=
λ1
6.63 × 10−34 J.s × 3 × 108 m/s
=
750 × 10−9 m
= 2.65 × 10−19 J.
Power of the light beam is 10−3 W = 10−3 J/s, i.e., 10−3 J energy is emitted per
second. Total number of photons emiitted will be,
Total energy
Energy of one photon
10−3 J
N1 =
2.65 × 10−19 J
= 3.77 × 1015 photons.
Number of photons =
Therefore K will be,
N1
I1
3.77 × 1015 photons
=
50 × 10−3 A
= 7.54 × 1016 photons/A.
K1 =
This constant of proportionality K1 is same for both the lights, therefore photoelectric
current due to second light will be,
I2 =
N2
.
K1
Energy of one photon for second light is,
E2 = hf2
hc
=
λ2
6.63 × 10−34 J.s × 3 × 108 m/s
=
400 × 10−9 m
= 4.97 × 10−19 J.
Spring semester 2015
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
Total number of photons emitted due to second light will be,
Total energy
Energy of one photon
10−3 J
N2 =
4.97 × 10−19 J
= 2.01 × 1015 photons.
Number of photons =
Therefore new photoelectric current will be,
N2
K1
2.01 × 1015 photons
=
7.54 × 1016 photons/A
= 0.0266 A
I2 =
= 26.6 mA.
Hence by decreasing the wavelength of light photoelectric current also decreases.
6. A P-N junction solar cell can run on light of maximum wavelength of 450 nm. We
provide it a forward bias and there is a current flowing through it. As the electrons
reach the P-type material, Or rather midway in the depletion region, they start to
annihilate the holes by filling them up. What minimum frequency of light will come
out as a result of this current conduction? Can these solar cells be used as sources of
light of desired frequency?
Answer 6
We are given that,
Maximum wavelength of light = λ = 450 nm = 450 × 10−9 m.
When electrons start to annihilate the holes, the minimum frequency of light that
would be emitted is the same as what was required to run, since the energy gap is
fixed. Therefore,
c
3 × 108 m/s
=
λ
450 × 10−9 m
= 6.67 × 1014 Hz.
f =
Yes, they can be used as source of light of desired frequency, if we carefully control
the voltage supplied to them. These will not be coherent sources of light though. The
frequencies would lie in some range rather than at a sharp value.
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
7. For a long time in medical history, X-rays have been considered to be a necessary evil,
though they are being replaced by other imaging techniques now. Explain why they
were necessary and why they were evil? Why light of longer wavelengths like visible
light or radio waves could not be used for imaging? Why their evil character can not
be done away by softening the beam?
Answer 7
X-rays are necessary because they provide us a way non-invasive diagnosis. We can
gauge the problems within the body without actually opening it up!
X-rays are evil because they are actually high energy photons and interact with the
metabolism at the cellular level. Also known to be carcinogenic.
Visible or radio waves have large wavelength and simply pass over cells without being
scattered appreciably. It is like the cells are invisible to them.
People are now actually using low intensity or low dosage of X-rays on patients but
each photon still has enough energy to cause damage. All it needs to do is one A to
G and you may be history (that is bit of an exaggeration though).
8. It is said that one should not get more than 10 exposures of normal intensity X-rays
in life time or one carries the risk of developing cancer. On the other hand we are
exposed to sunlight all the time. Which exposure carries more energy? Why sunlight
is not dangerous? It is claimed by many people that cell phone signals can cause
cancer. What is your educated guess about it based on what you have learnt about
light so far. Remember, cell phone signals work in radio frequencies.
Answer 8
Sunlight has more energy since our exposure time is greater and there are all sorts of
electromagnetic waves from the spectrum to which we are exposed. Sunlight is not
dangerous as compared to X-rays since it mostly contains visible light, infrared and
other longer wavelengths. These do not interact much at the cellular level.
No, cell phones cannot cause cancer (at least not because of the radio waves). The
waves have simply too large wavelength to interact.
9. X-rays of wavelength 0.200 nm are scattered from a block of carbon. If the scattered
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
radiation is detected at 90◦ to the incident beam, find the shift in the beams wavelength
and also find the kinetic energy imparted to the recoiling electron. At what angle we
will find the beam with greatest shift in wavelength?
Answer 9
We are given that,
Wavelength of X-rays = λ = 0.200 nm = 0.200 × 10−9 m = 2 × 10−10 m
Angle of scattered photon = θ = 90◦ .
We need to calculate,
Wavelength shift = ∆λ =?
Kinetic energy of recoiling electron = Ee =?
Angle of greatest wave shift = θ =?
Wavelength shift can be calculated by using the following relation,
h
1 − cos θ
me c
6.63 × 10−34 m2 kg.s−1
=
(1 − cos 90◦ )
9.1 × 10−31 kg × 3 × 108 m/s
= 2.4 × 10−12 m = 2.4 × 10−3 nm.
∆λ =
Kinetic energy imparted to the electron will be equal to the energy lost by the photon,
Energy imparted to electron = Energy of incident photon − Energy of scattered photon
Ee = E − E 0
hc hc
=
−
λ λ0 1
1
= hc
− 0 ,
λ λ
where λ0 = ∆λ + λ = 2.4 × 10−12 m + 2 × 10−10 m = 2.024 × 10−10 m. Thus kinetic
energy of photon will be,
−34
Ee = 6.63 × 10
2
−1
m kg.s
= 1.18 × 10−17 J =
× 3 × 10 m/s
8
1
1
−
−10
2 × 10
m 2.024 × 10−10 m
1.18 × 10−17
eV
1.6 × 10−19
= 73.7 eV.
Spring semester 2015
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
In order to calculate angle for which wavelength shift is maximum, let’s differentiate
∆λ with respect to θ and equate it to zero.
d
d
h
∆λ =
1 − cos θ
dθ
dθ me c
d
h
h
·
1 − cos θ =
· 0 − (− sin θ) .
=
me c dθ
me c
Equate last equation to zero.
h
· 0 − (− sin θ) = 0
me c
h
sin θ = 0
me c
⇒ sin θ = 0
θ = π.
Hence if scattered angle θ = π, the beam will scatter with greatest wavelength shift.
This is called backscattering.
10. Photons of wavelength 0.0711 nm are bombarded on a crystal. What is the wavelength
of backscattered photons, the ones that come right back? What is the energy of these
photons? What is the energy of the recoiled electron?
Answer 10
We are given that,
Wavelength of photon = λ = 0.0711 nm = 0.0711 × 10−9 m = 7.11 × 10−11 m
Angle of scattered photon = θ = π.
we need to calculate,
wavelength of backscattered photon = λ0 =?
Energy of backscattered photon = E 0 =?
Energy of recoiled electron = Ee =?
Spring semester 2015
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
According to Compton effect, shift in wavelength is given by,
∆λ = λ0 − λ =
λ0 = λ +
h
me c
h
1 − cos θ
me c
1 − cos θ
6.63 × 10−34 m2 kg.s−1
(1 − cos π)
9.1 × 10−31 kg × 3 × 108 m/s
m = 0.076 nm.
= 7.11 × 10−11 m +
= 7.5957 × 10−11
Energy of backscattered photon will be,
hc
λ0
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
=
7.5957 × 10−11 m
2.6 × 10−15
= 2.6 × 10−15 J =
eV
1.8 × 10−19
= 16366 eV = 16.4 keV.
E0 =
Energy of recoil electron will be,
hc
−E0
λ
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
− 2.6 × 10−15 J
−11
7.11 × 10
m
2.797 × 10−15 J − 2.6 × 10−15 J
1.97 × 10−16
1.97 × 10−16 J =
eV
1.6 × 10−19
1234 eV = 1.23 keV.
Ee = E − E 0 =
=
=
=
=
11. In a Compton scattering, the scattered photon has an energy of 120 keV and the
recoiled electron has energy of 40 keV . Find the wavelength of the incident photon as
well as the angles at which both electron and the photon scatter relative to the initial
direction of the photon.
Answer 11
We are given that,
Energy of scattered photon = E 0 = 120 keV
Energy of recoiled electron = Ee = 40 keV.
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
We need to calculate,
Wavelength of incident photon = λ =?
Angle of recoiled electron = φ =?
Angle of scattered photon = θ =?
Wavelength of incident photon can be calculated by using the equation,
hc
λ
hc
⇒λ =
,
E
E =
where according to law of conservation of energy,
Total energy before collision = Total energy after collision
Energy of incident photon = Energy of recoiled electron + Energy of scattered photon
E = Ee + E 0
= 40 keV + 120 keV
= 160 keV = 160 × 103 eV
= 160 × 103 × 1.6 × 10−19 J
= 2.56 × 10−14 J.
Therefore wavelength of incident photon will be,
hc
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
=
E
2.56 × 10−14 J
= 7.76 × 10−12 m = 7.76 × 10−3 nm.
λ =
Angle of scattered photon θ can be calculated by using the following relation.
h
∆λ = λ0 − λ =
1 − cos θ
me c
me c 0
−1
⇒ θ = cos
1−
λ −λ ,
h
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
where λ0 can be calculated by using the energy equation of scattered photon as follows.
hc
λ0
hc
⇒ λ0 =
E0
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
=
120 keV
−34
6.63 × 10
m2 kg.s−1 × 3 × 108 m/s
=
120 × 103 × 1.6 × 10−19 J
= 1.0 × 10−11 m.
E0 =
Therefore angle of scattered photon will be,
9.1 × 10−31 kg × 3 × 108 m/s
−1
−11
−12
1−
θ = cos
1.0 × 10
m − 7.76 × 10
m
6.63 × 10−34 m2 kg.s−1
= 94◦ .
Hence scattered photon will have an angle of 94◦ with the initial direction of incident
photon.
Angle of recoiled electron can be calculated by using law of conservation of momentum
along y-axis, according to which,
Total momentum after collision 
 Total momentum before collision
 = 
 Momentum of scattered photon 
 Momentum of incident photon 






 = 
+
+




Momentum of recoiled electron
Momentum of electron
0 + 0 = pe sin φ + (−p 0 sin θ)
0 = pe sin φ − p 0 sin θ
pe sin φ = p 0 sin θ
0
−1 p
φ = sin
sin θ ,
pe
where
p
p
2me Ee = 2 × 9.1 × 10−31 kg × 40 keV
p
= 2 × 9.1 × 10−31 kg × 40 × 103 × 1.6 × 10−19 J
pe =
and
Spring semester 2015
= 1.1 × 10−22 kgm/s
h
6.63 × 10−34 kgm2 s−1
0
p = 0 =
λ
1.0 × 10−11 m
= 6.63 × 10−23 kgm/s.
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Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
Therefore angle of scattered photon will be,
−23
kgm/s
−1 6.63 × 10
◦
φ = sin
× sin 94
1.1 × 10−22 kgm/s
= 37◦ .
Hence recoiled electron will have an angle of 37◦ with the initial direction of incident
photon.
12. You want to map the structure of a crystal of inter molecular spacing of 5 nm. Which
one of these photons can do the best job in a diffraction experiment? photons of (a)
250 eV (b) 5 eV (c) 100 keV (d) All of them can do the job equally well as we are
using individual photons which are particles.
Answer 12
We are given that,
Intermolecular spacing = d = 5 nm
(a) Energy of photon = E1 = 250 eV = 250 × 1.6 × 10−19 J = 4 × 10−17 J
(b) Energy of photon = E2 = 5 eV = 5 × 1.6 × 10−19 J = 8 × 10−19 J
(c) Energy of photon = E3 = 100 keV = 100 × 103 × 1.6 × 10−19 J = 1.6 × 10−14 J
(d) All of them can do the job equally well as we are using individual photons which
are particles.
In order to investigate the inner structure of a crystal, wavelength of the light used
should be comparable with the intermolecular spacing, so option (d) can never be a
true option. Let’s calculate the wavelength of waves by using three different energies.
λ1 =
=
λ2 =
=
λ3 =
=
Spring semester 2015
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
hc
=
E1
4 × 10−17 J
4.97 × 10−9 m = 4.97 nm
hc
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
=
E2
8 × 10−19 J
2.49 × 10−7 m = 249 nm
hc
6.63 × 10−34 m2 kg.s−1 × 3 × 108 m/s
=
E3
1.6 × 10−14 J
1.24 × 10−11 m = 0.0124 nm.
17
Modern Physics: Solution to Home work 2
Due date: 20 Feb. 2015
From above we can see that the wavelength λ1 = 4.97 nm is comparable to the
interatomic spacing d = 5 nm. Therefore photons of 250 eV are best to do the job in
a diffraction experiment.
Spring semester 2015
18