Phys 53 – S09
Review Exercise - Mid-Term 1
Dr. Ray Kwok
1) A 3.00-W beam of light of wavelength 125 nm falls on a metal surface. You observe that the
maximum kinetic energy of the ejected electrons is 4.20 eV. Assume that each photon in the
beam ejects a photoelectron.
(a) What is the work function (in electron volts) of this metal?
(b) How many photoelectrons are ejected each second from this metal?
(c) If the power of the light beam, but not its wavelength, were reduced by half, what would
be the answer to part b?
(d) If the wavelength of the beam, but not its power, were reduced by half, what would be the
answer to part b?
(a)
for photoelectrons, max kinetic energy Kmax = hc/λ − φ
φ = hc/λ – Kmax = (1.244 eV-µm)/(0.125 µm) – 4.20 eV = 5.75 eV
(b)
energy of a photon = hc/λ = (1.244 eV-µm)/(0.125 µm) = 9.95 eV
9.95 eV (1.6 x 10-19 J / eV) = 1.59 x 10-18 J per photon
N = (3 J/s)(1 photon / 1.59 x 10-18 J ) = 1.88 × 1018 electrons/s
(c)
Power reduced by half, number of photons emitted reduced by half
N = 9.4 × 1017 electrons/s
(d)
If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ. To
maintain the same power, the number of photons must be half of what they were in part
(b), so N is cut in half to 9.4 × 1017 electrons/s. We could also see this from part (b),
where N is proportional to λ. So if the wavelength is cut in half, so is N.
Phys 53 – S09
Review Exercise - Mid-Term 1
Dr. Ray Kwok
2) In high-energy physics, new particles can be created by collisions of fast-moving projectile
particles with stationary particles. Some of the kinetic energy of the incident particle is used to
create the mass of the new particle. A proton-proton collision can result in the creation of a
negative kaon K− and a positive kaon K+
p + p → p + p + K− + K+
(a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to
occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7
MeV, and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in
the frame in which the total momentum is zero – rest frame. Note that here the Lorentz
transformation must be used to relate the velocities in the laboratory frame to those in the
rest frame.)
(b) How does this calculated minimum kinetic energy compare with the total rest mass energy
of the created kaons?
(c) Suppose that instead the two protons are both in motion with velocities of equal magnitude
and opposite direction. Find the minimum combined kinetic energy of the two protons that
will allow the reaction to occur.
(d) How does this calculated minimum kinetic energy compare with the total rest mass energy
of the created kaons? (This example shows that when colliding beams of particles are used
instead of a stationary target, the energy requirements for producing new particles are
reduced substantially.)
In the center-of-mass frame (rest frame), the 2 protons have the same speed and KE. The minimum
KE in this frame is just equal to the rest mass of kaons. So, KE = 493.7 MeV for each proton.
KE = (γ – 1)moc2
493.7 = (γ – 1)(938.3), or γ = 1.526
(v/c) 2 = 1 – 1/γ2 or
v = 0.755 c (this is
the speed of each proton in the rest frame)
y’
u = 0.755c
z’
proton
v = - 0.755c
x’
y
proton
CM
x
z
(a) In a proton’s reference frame, the other
proton is moving at a speed of:
v’ = [v – u]/[1 – vu/c2]
= [– 0.755 – 0.755]c/[1 + (0.755)2]
= − 0.962 c
γ = 1/√[1 – (v/c)2] = 1/√[1 – (0.962)2]
γ = 3.65
KE = (γ – 1)moc2 = (3.65 – 1)(938.3)
KE = 2488 MeV
(b) KE/2mkc2 = 2488/2(493.7) = 3.65
(c)
This is the same as the min KE in the rest frame. KE = 493.7 MeV for each proton.
Total KE needed is 2(493.7) = 987.4 MeV
(d)
total KE/2mkc2 = 987.4/2(493.7) = 1
so it requires much less energy to collide beams of particles than using stationary targets.
Phys 53 – S09
Review Exercise - Mid-Term 1
Dr. Ray Kwok
3) Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they
stand 24.0 m from each other. Being very skilled players, they play without a net. The tennis
ball has mass 0.0580 kg. You can ignore gravity and assume that the ball travels parallel to the
ground as it travels between the two players. Unless otherwise specified, all measurements are
made by the two men.
(a) Lorentz serves the ball at 77.0 m/s . What is the ball's kinetic energy?
(b) Einstein slams a return at 1.76×108 m/s. What is the ball's kinetic energy?
(c) During Einstein's return of the ball in part A, a white rabbit runs beside the court in the
direction from Einstein to Lorentz. The rabbit has a speed of 2.21×108 m/s relative to the
two men. What is the speed of the rabbit relative to the ball?
(d) What does the rabbit measure as the distance from Einstein to Lorentz?
(e) How much time does it take for the rabbit to run 24.0 m, according to the players?
(f) The white rabbit carries a pocket watch. He uses this watch to measure the time (as he sees
it) for the distance from Einstein to Lorentz to pass by under him. What time does he
measure?
(a)
KE = ½ mov2 = ½ (0.058 kg)(77 m/s)2 = 172 J
(b)
v = 1.76×108 m/s = 0.587 c
γ = 1/√[1 – (v/c)2] = 1/√[1 – (0.587)2] = 1.235
KE = (γ – 1)moc2 = (1.235 – 1)(0.058 kg)(3 x 108 m/s)2 = 1.23 x 1015 J
(c)
rabbit v = 2.21×108 m/s = 0.737c relative to the players.
y’
u = 0.587c
y
tennis ball
z’
x’
v = 0.737c
v’ = [v – u]/[1 – vu/c2]
= [0.737 – 0.587]c/[1 − (0.737)(0.587)]
v’ = 0.264 c is the velocity of Rabbit
relative to the tennis ball.
rabbit
players
x
z
(e)
∆t = ∆Lo/v = 24/2.21 x 108 = 109 ns
(f)
∆t’ = ∆L/v = 16.2/2.21 x 108 = 73.4 ns
check: ∆t = γ∆to = (1.48)(73.4) = 109 ns
(d) γ = 1/√[1 – (v/c)2]
= 1/√[1 – (0.737)2] = 1.48
∆L = ∆Lo/γ = 24/1.48 = 16.2 m
Phys 53 – S09
Review Exercise - Mid-Term 1
Dr. Ray Kwok
4) The negative muon has a charge equal to that of an electron but a mass of 207 times as great.
Consider a hydrogen-like atom consisting of a proton and muon. (a) What is the reduced mass
of the atom? (b) What is the ground-level energy in eV? (c) What is the wavelength of the
radiation emitted in the transition from the n = 2 level to the n = 1 level?
(a) Reduced mass mr = m1m2/(m1 + m2) = 207memp/(207me + mp) = 1.69 x 10-28 kg
(b) new energy levels are: (using equation of energy levels for Bohr hydrogen-like atoms)
En = mre4/[εo2 8n2h2] = mr/me [−13.6/n2]
for ground state (n=1)
E1 = −2.53 keV
(c) hc/λ = E2 – E1
hc = (4.15 x 10-15 eV-s)(3 x 108 m/s) = 1.244 eV-µm
En = −2.53/ n2 keV
E2 – E1 = − 2.53(1/22 – 1) = 1.9 keV
λ = hc/(E2 – E1) = (1.244 eV-µm)/(1.9 keV) = 6.55 Å