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PC 242 Assignment 2: Solutions
1. An electron has a momentum that is 90% larger than its predicted classical momentum.
How fast is it going? Calculate its total energy and kinetic energy in MeV. How would
your answers change if the particle were a proton?
The electron momentum is
p = γ mv = 1.90mv =
mv
$1 − ( v c )2 '
)(
%&
12
⇒
1
12
$1 − (v c) 2 '
)(
%&
$ + 1 .2'
= 1.90 ⇒ v = &1 − 0 )
&% , 1.90 / )(
12
c
= 0.85c
Total energy of the electron is
E = γ me c 2 = 1.9 × 0.511MeV = 0.97MeV
Kinetic energy of the electron = total energy – rest energy
KE = γ me c 2 − me c 2 = 0.46MeV
For a proton,
p = γ mv = 1.90mv =
mv
$1 − ( v c )2 '
&%
)(
12
⇒
1
12
$1 − (v c) 2 '
&%
)(
$ + 1 .2'
= 1.90 ⇒ v = &1 − 0 )
&% , 1.90 / )(
12
c
= 0.85c
Notice that the mass canceled out, yielding the same answer as for the electron.
Total energy of the proton is
E = γ m p c 2 = 1.9 × 938.3MeV = 1783MeV
Kinetic energy of the proton = total energy – rest energy
KE = γ m p c 2 − m p c 2 = 844MeV
2. A 16kg object moving to the right at 0.6c collides with a 9kg object moving to the left
at 0.8c. After the collision the 16kg moves to the left at 0.6c.
(a) What is the velocity of the 9kg object?
(b) Verify that momentum is conserved in a frame moving to the right at 0.6c.
Conservation of momentum: Total momentum before collision = total momentum after
collision
γ 1m1u1 + γ 2 m2u2 = γ 3 m1u3 + γ 4 m2u4
m1 = 16kg, m2 = 9kg,u1 = 0.6c,u2 = −0.8c,u3 = −0.6c
Solving for u4 in the above equation gives a velocity of 0.8c to the right for the 9kg
object.
In a frame S’ moving to the right with v=0.6c conservation of momentum should still
hold,
Using the Lorentz transformations to calculate the velocities in the moving frame,
u" =
u −v
1 − uv c
2
with m1 = 16kg, m2 = 9kg,u1 = 0.6c,u2 = −0.8c,u3 = −0.6c,u4 = 0.8c
it is simple to verify that
γ 1" m1u1" + γ 2" m2u2" = γ 3" m1u3" + γ 4" m2u4"
3. The nucleus of a beryllium atom has a mass of 8.0031 amu, where amu is an atomic
mass unit: 1 amu = 1.66 x 10-27kg. It spontaneously breaks up into two identical pieces
each of mass 4.0015 amu. Assuming the nucleaus to be initially at rest, at what speed
will the fragments move after the fission and how much kinetic energy is released?
Conservation of energy: Total energy before decay = total energy after decay
γ Mc 2 = γ 1m1c 2 + γ 2 m2 c 2
Since the Beryllium atom is at rest, γ = 1. Also since m1=m2 the two daughter particles
will have equal and opposite velocities. Therefore γ1=γ2.
Hence the above equation becomes
Mc 2 = 2γ 1m1c 2
1
M
⇒γ1 =
=
v 2 2m1
1 − 12
c
2
v
4m12
⇒ 1 − 12 =
c
M2
v1 = 1 −
4m12
c = 1.5 × 10 6 m / s
2
M
The total energy after decay = total rest energy + kinetic energy
⇒ 2γ 1m1c 2 = 2m1c 2 + KE
⇒ KE = 2(γ 1 − 1)m1c 2 = 1.5 × 10 −14 J = 93.8keV
4. A light source of wavelength λ illuminates a metal and ejects electrons with a
maximum kinetic energy of 1.00 eV (electron-volt). A second light source with half the
wavelength of the first ejects electrons with a maximum kinetic energy of 4.00 eV. What
is the work function of the metal? What is the maximum wavelength of light that will
cause electrons to be ejected from the plate?
Equation describing the photoelectric effect:
Max KE of electrons = incoming photon energy–work function (binding energy) of metal
hc
hc
K max = hf − W =
−W ⇒W =
− K max ;
λ
λ
hc
First Source: W =
− 1.00 eV .
λ
hc
2hc
Second Source: W = λ − 4.00 eV=
− 4.00 eV .
λ
2
As the work function is the same for both sources (a property of the metal),
hc
2hc
hc
− 100 eV =
− 4.00 eV ⇒ = 3.00 eV and
λ
λ
λ
W=
hc
− 1.00 eV = 3.00 eV − 1.00 eV = 2.00 eV .
λ
The maximum wavelength of light that will cause electrons to be ejected is the
wavelength at which the photon has just enough energy to overcome the work function
W:
hc
hc
=W ⇒λ =
= 620nm
λ
W
5. A photon and an electron undergo Compton scattering. After the collision, the electron
has an energy of 36keV. The scattered photon has an energy of 115keV. What is the
frequency of the incident photon, and the scattering angles of the photon and electron?
From conservation of energy,
total energy of electron and photon before collision = total energy of electron + photon
after collision . Therefore,
E 0 + me c 2 = E ! + me c 2 + K e
where E is the energy if the incoming photon, E’ is the energy of the outgoing photon and
Ke is the kinetic energy of the scattered electron. Thus
0
E0 = E ! + K e = 115 keV + 36 keV=151 keV .
The incoming photon energy can be written as E0 =
hc
. This gives
λ0
€
λ0 =
(b)
hc 1 240 nm eV
=
= 8.21 × 10 −3 nm .
3
E0 151 × 10 eV
Using the Compton scattering relation λ " − λ0 = λ c (1 − cos θ ) where
hc 1 240 nm eV
h
= 0.002 43 nm and λ ! =
=
= 10.8 × 10 −3 nm .
me c
E ! 115 × 10 3 eV
Solving the Compton equation for cos θ , we find
λc =
− λc cosθ = λ $ − λ0 − λc
λ $ − λ0
cosθ = 1 −
= −0.066
λc
The angle is
(c)
θ = 93.8° .
Using the conservation of momentum equations (see Compton scattering notes)
one can solve for the recoil angle of the electron.
p = p" cos θ + pe cos φ
pe sin φ = p# sin θ ;
dividing these equations one can solve for the recoil angle of the electron
h
sin θ
p" sin θ
& hc )
λ
"
tan φ =
= h h
=( +
p − p" cosθ λ0 − λ " cosθ ' λ " *
=
€
and
φ = 36° .
hc
λ0
sin θ
− λ " hc
cos θ
115 keV ( 0.9978 8 )
114.7562
=
= 0.723
151 keV − 115 keV ( −0.066 )
158.59