Chapter 3
Physics 205 Solution of Home Work
Problems
3.1
Problem 3.10
(a) Use Stefan’s law to calculate the total power radiated per unit area by a tungsten filament
at a temperature of 3000 K. (Assume that the filament is an ideal radiator.)
(b) If the tungsten filament of a light bulb is rated at 75 W , what is the surface area of the
filament? (Assume that the main energy loss is due to radiation.)
Solution
(a) Stefan’s law (see equation 2.3 of the text book) states:
e = σT 4
where e is the radiated power per unit area, σ is the Stefan-Boltzman constant =
5.6705×10−8 W/m2 · K 4 and T is absolute temperature. let assume that the total
power radiated is P and the surface area of the filament is A; we then have:
P
A
= σT 4
= 5.6705 × 10−8 × (3000)4
= 4.62 × 106 W/m2
e =
2
CHAPTER 3. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
(b) The surface area of the filament that is rated at 75 W , is given by:
A =
P
e
75
4.62 × 106
= 1.62 × 10−5 m2
= 16.2 mm2
=
3.2. PROBLEM 3.23
3.2
3
Problem 3.23
The average power generated by the sun has the value 3.74 × 1026 W . Assuming the average
wavelength of the Sun’s radiation to be 500 nm, find the number of photons emitted by the
sun in 1 s.
Solution
Each photon carries an amount of energy of hν = hc/λ, where h is Planck’s constant =
6.63×10−34 J · s, ν is the frequency and λ the wavelength of the photon’s radiation. The
power P radiated by the Sun is 3.74×1026 W or 3.74×1026 J/s at an average wavelength of
λ = 500 nm, the number of photons emitted in 1 s, n/t is then:
n
P
=
photon/s
t
hν
Pλ
=
hc
3.74 × 1026 × 500 × 10−9
=
6.63 × 10−34 × 3 × 108
= 9.40 × 1044 photon/s
4
3.3
CHAPTER 3. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
Problem 3.28
A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum
kinetic energy of 1.00 eV . A second light source with half the wavelength of the first ejects
photoelectrons with a maximum kinetic energy of 4.00 eV . What is the work function of the
metal?
Solution
The maximum kinetic energy Kmax is given by:
Kmax = hν − φ
hc
=
−φ
λ
we then have, for the two cases:
hc
−φ
λ1
hc
=
−φ
λ2
K1max =
K2max
Since λ2 = 12 λ1 and the metal is the same we then get:
hc
− K1max
λ1
hc
=
− K2max
λ2
hc
= 2 − K2max
λ1
φ =
(3.1)
(3.2)
Using Equations (2.1) and (2.2) we get:
hc
hc
− K1max = 2 − K2max
λ1
λ1
or
K2max − K1max =
and
λ1 =
hc
λ1
hc
K2max − K1max
(3.3)
3.3. PROBLEM 3.28
5
Substitution for λ1 from Equation (2.3) into equation (2.1) we get:
φ =
hc
hc
K2max −K1max
− K1max
= K2max − 2K1max
= 4.0 − 2.0
= 2.0 eV
6
CHAPTER 3. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
3.4
Problem 3.37
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered
rays are detected at 30◦ relative to the incident rays, find:
(a) The compton shift at this angle,
(b) The energy of the scattered X-rays, and
(c) The energy of the recoiling electron
Solution
(a) The Compton shift ∆λ is given by:
∆λ = λ0 − λ◦
h
(1 − cos θ)
=
me c
hc
(1 − cos θ)
=
me c2
1.240 × 103 (eV · nm)
=
(1 − cos 30◦ )
511 × 103 (eV )
= 3.25 × 10−4 nm
Where λ◦ is the wave of the incident X-rays and λ0 is the wave of the scattered X-rays.
(b) We calculate the wavelength λ0 of the scattered x-rays as:
E◦ =
λ◦ =
λ0 =
=
=
=
=
hc
λ◦
hc
E◦
λ◦ + ∆λ
hc
+ ∆λ
E◦
1.240 × 103 (eV · nm)
+ 3.25 × 10−4 (nm)
3 × 105 (eV )
4.458 × 10−3 nm
4.458 × 10−12 m
3.4. PROBLEM 3.37
7
The energy E 0 of the scattered X-rays is:
hc
λ0
(4.14 × 10−15 eV · s)(3 × 108 m/s)
=
4.458 × 10−12 m
= 2.79 × 105 eV
= 279 keV
E0 =
(c) The recoiling electron energy Ee can be calculated from consideration of energy conservation.
E◦ = E 0 + Ke
Ke = E◦ − E 0
= 300 − 279
= 21 keV
8
CHAPTER 3. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
3.5
Problem 3.43
◦
In a Compton collision with an electron, a photon of violet light (λ = 4000 A) is back
scattered through an angle of 180◦ .
(a) How much energy (eV ) transferred to the electron in this collision?
(b) Compare your result with the energy this electron would acquire in a photoelectric
process with the same photon.
(c) Could violet light eject electrons from a metal by Compton collision? Explain.
Solution
(a) The energy ∆Ee transferred to the electron can be calculated from the Compton shift,
i.e.
∆λ = λ0 − λ◦
hc
=
(1 − cos θ)
me c2
hc
= 2
me c2
1.240 × 103 (eV · nm)
= 2
511 × 103 eV
= 4.85 × 10−3 nm
Since
E=
hc
λ
then
hc
∆E = −
∆λ
λ2
1.240 × 103 (eV · nm)
∆Ee =
× 4.85 × 10−3 nm
2
2
(400) nm
= 3.76 × 10−5 eV
3.5. PROBLEM 3.43
9
(b) In photoelectric effect, the photon give all its energy to the electron, i.e.
hc
λ
1.240 × 103 (eV · nm)
=
400 nm
= 3.1 eV
Ephoto =
Comparing Compton scattering energy transfer to photoelectric energy transfer we get:
3.76 × 10−5
∆Ee
=
Ephoto
3.1
= 1.21 × 10−5
(c) By comparing the results of Problem (2.19) and Problem (2.37a), we conclude that the
answer is no. Problem (2.19) shows that a typical work-function of a metal is about 2
eV while Problem (2.37a) shows that the maximum energy given to an electron by an
ultraviolet photon in Compton scattering is 3.76 × 10−5 eV