# Physics 2 practice exam ```Recitation 9
Chapter 28
Problem 28.14. Calculate the energy and momentum of a photon of wavelength 700 nm.
You should be familiar with these equations by now (after our time with the Bohr atom and relativity). The energy is given
by (Eq’s 28.3 and 11.15)
1240 eV&middot;nm
hc
=
= 1.77 eV .
(1)
E=
λ
700 nm
For momentum you can either use the relativistic energy-momentum equation (Eq. 9.22)
E 2 = p2 c2 + m20 c4
Ephoton = pc
E
p=
= 1.77 eV/c = 9.46 &middot; 10−28 kg&middot;m/s
c
(2)
(3)
(4)
or the de Broglie formula (Eq. 28.10)
p=
6.63 &middot; 10−34 J&middot;s
h
=
= 9.46 &middot; 10−28 kg&middot;m/s .
λ
700 nm
(5)
Problem 28.15. X-rays having an energy of 300 keV undergo Compton scattering from a target. The scattered rays are
detected at 37.0◦ relative to the incident rays. Find (a) the Compton shift at this angle, (b) the energy of the scattered x-ray,
and (c) the energy of the recoiling electron.
(a) From the Compton shift equation (Eq. 28.8)
h
(1 − cos θ)
me c
∆λ = 2.43 pm(1 − cos 37.0◦ ) = 489 fm .
λ0 − λ0 =
(6)
(7)
(b) The wavelength of the incoming photon was
λ0 =
hc
1240 eV&middot;nm
= 4.13 pm .
=
E0
300 keV
(8)
The scattered wavelength is thus
λ0 = λ0 + ∆λ = (4.13 + 0.489) pm = 4.62 pm ,
(9)
and the energy of the scattered photon is
E0 =
hc
1240 eV&middot;nm
=
= 268 keV .
0
λ
4.62 pm
(10)
(c) All the energy lost by the photon must go into the recoiling electron so
Ee = E0 − E 0 = (300 − 268) keV = 31.7 keV .
(11)
Problem 28.16. A 0.110 nm photon collides with a stationary electron. After the collision, the electron moves forward and
the photon recoils backward. Find the momentum and the kinetic energy of the electron.
The photon scatters by 180◦ , so from the Compton shift equation
λ0 = λ0 +
h
2h
(1 − cos 180◦ ) = λ0 +
= (110 + 4.85) pm = 115 pm .
me c
me c
(12)
The kinetic energy of the electron is given by the change in photon energy (just like problem 28.15).
K = E0 − E 0 =
hc hc
− 0 = 11.3 keV − 10.8 keV = 476 eV .
λ0
λ
(13)
We conserve momentum to find the electron’s momentum, using pphoton = E/c.
E0
E0
= pf = pe −
c
c
E0 + E 0
pe =
= (11.3 + 10.8) keV/c = 22.1 keV/c = 1.18 &middot; 10−23 kg&middot;m/s .
c
pi =
(14)
(15)
```