PHYS 203 Homework #5

PHYS 203 Homework #5
Solutions
1. (5 pt.) A positron is an anti-electron. It has the same mass but opposite charge (and
some other “opposites.”) If an electron and a positron collide, they annihilate and the
mass disappears, replaced by two gamma rays. Why two It’s because momentum
must be conserved. In the rest frame of the electron-positron pair’s center of mass, the
momentum is zero. If only one gamma ray were produced, the momentum could not
be zero, since all photons (like gamma rays) have momentum equal to E/c. But two
gamma rays, traveling in opposite directions, can keep the total momentum equal to
zero.
Suppose an electron and a positron are both traveling at 0.450c in opposite directions,
as measured by someone in the lab frame. They have a head-on collision and annihilate,
producing two gamma rays. What is the energy of each gamma ray (You don’t have
to know anything about gamma rays. Just assume the total energy is split between
them.) Express the energy in MeV. For an electron, mc 2 = 0.511 MeV.
The total energy of each particle is
E =
mc 2
√
1 − v2 /c 2 =
= (0.511 MeV)(1.120)
= 0.572 MeV
0.511 MeV
√
1 − (0.450)
2
The total energy is just twice this. But this energy is split between the two gamma
rays, so the energy of each gamma ray is simply the energy we found above.
2. (5 pt.) In the previous problem, each particle was moving at 0.450c as measured in the
lab frame, and they were approaching each other. How fast is the positron traveling
as viewed from the electron’s frame of reference Be careful in setting this up. One
of the velocities you need is that of the lab relative to the electron, not the electron
relative to the lab.
0.450 c
−0.450 c
S S’
v = 0.450c
(positron)
Earth frame
(electron)
u’
Suppose the positron is traveling to the right, and the electron to the left, in the Earth
(lab) frame. We need to transform to one of the particle frames, so let us choose the
electron frame. The Earth is traveling to the right at speed v = 0.450c, when we are
in the electron frame. In the diagram, we call the electron frame S, the Earth frame

S ′ , and the speed of the positron is u ′ in S ′ . We can now apply the velocity addition
relation:
u = v + u′
1 + vu′
c 2
0.450c + 0.450c
=
1 + (0.450) 2
= 0.75c
3. (4 pt.) Suppose a rocket ship is approaching Earth at a speed of 0.80c. The rocket
sends out a space probe toward Earth, launching it at a speed of 0.40c relative to the
rocket. As measured by someone on Earth, how fast is the probe traveling
Probe
0.40c
S S’
v = 0.80c
Rocket frame
(sitting still)
0.80c
Earth
u’
In the diagram (left side), the we are in the rocket frame, with the Earth traveling left
at 0.80c and the probe traveling to the right at 0.40c. We then transform to the Earth
frame. On the right side of the diagram, the Earth frame is S and the rocket frame is
S ′ . The speed of the probe in S ′ is u ′ = 0.40c. We apply the velocity addition relation:
u = v + u′
1 + vu′
c 2
0.80c + 0.40c
=
1 + (0.80)(0.40)
= 0.91c
4. (6 pt.) In a Young’s double-slit measurement, suppose that a helium-neon laser is used
for the light source. The light has a wavelength of 633 nm. We measure the 8th-order
bright fringe at 10.0 ◦ from the normal.
(A) What is the separation between the two slits (Answer: 29.2 µm)
We use the double-slit equation:
d sin θ = mλ
d sin 10.0 ◦ = (8)(0.633 µm)
5.064 µm
d = = 29.1 µm
0.1736
(B) Suppose we repeat the experiment, but this time we fill a fish tank with water
and submerge the double slit in the water. When the laser light passes through the

slit, at what angle (from the normal) will the 8th-order bright fringe be seen (Recall:
λ ′ = λ/n).
The wavelength is now smaller, so the angle will be less:
d sin θ = m λ n
(8)(0.633 µm)
sin θ =
(29.1 µm)(1.33) = 0.131
θ = 7.5 ◦
5. (5 pt.) In the lab we have a diffraction grating which is labeled “600 lines/mm.” (A)
Determine the separation between the lines on the grating, in µm.
1 mm is 1000 µm, so to find the spacing in µm, we divide 1000 µm by 600:
d =
1000 µm
600
= 1.667 µm
(B) If we shine light from an Argon-ion laser, with λ = 488 nm, on the grating at
normal incidence, at what angle from the normal will each of the first three orders (m
= 1, 2, and 3) of bright fringes occur
We use the double-slit interference relation:
d sin θ = mλ
sin θ = mλ
d
(1)(0.488 µm)
=
1.667 µm = 0.2927
θ = 17.0 ◦ for m = 1
For m = 2, the value of sinθ is twice 0.2927:
θ = arcsin 2(0.2927) = 35.8 ◦ for m = 2
Similarly, for m = 3, the value of sinθ is three times 0.2927:
θ = arcsin 3(0.2927) = 61.4 ◦ for m = 3
(C) Will there be any other orders of constructive interference
No, there will be no more orders of contructive interference, because 4(0.2927) is greater
than 1, so there is no angle that will work for this. Another way to say it is, 4 times
the wavelength is greater than the separation between slits (or scratches), so even at
90 ◦ there cannot be a 4-wavelength difference between the path lengths of the light
from adjacent slits.