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PHGN310 Modern Physics Final Exam May 2, 2002
- (15) In the movie K-PAX the protaganist, Prot, claims to travel at tachyonic (faster that light) speeds to get from his planet, 1000 light-years away, to earth. Since we don’t have tachyonic technology yet, humans decide to make the journey using a rocket ship which is capable of traveling at 0.9999 c. (a) (5) How long would the trip to K-PAX take as viewed from an observer who stays on earth?
From earth’s viewpoint the time is just the distance divided by the speed: tE = d 0 /v.
(b) (5) How long does the trip take according to the space ship’s passengers?
There are two ways to look at this problem: from the viewpoint of the traveler the distance is contracted by the inverse gamma-factor so the time of travel is t = (d 0 /γ)/v; alternatively, according to an earth observer the clocks on the spaceship run slower so the time is: tT = (d 0 /v)/γ which gives the same result.
(c) (5) The gross annual energy production of the United States in 2000 was about 10^20 J. Suppose the spaceship weighs 10,000 kg. How many years worth of U.S. energy production is needed to have the enough energy to get the spaceship to a speed of 0.9999 c?
The kinetic energy of the spaceship is: KE = mc^2 (γ − 1). Dividing this by the annual energy production gives the number of years worth of energy needed to get the ship to this speed.
- (15) (a) (5) Neutrons absorbed by nuclei typically release a gamma ray equal to the binding energy of that captured neutron. What is the energy of the gamma ray emitted when a neutron is captured on 40 Ca? (DATA: M(n) = 1.008665 u, M(^40 Ca)= 39.962591 u, M(^41 Ca)= 40.962278u, u=931.5 MeV/c^2 )
Eγ = (M [^40 Ca] + M [n] − M [^41 Ca])c^2. (Because of the heavy mass of the 41 Ca nucleus, one can ignore recoil.)
(b) (5) How much energy is released in the reaction: 2 H + 3 H → 4 He + n? (DATA: M (^2 H) = 2.014102 u; M (^3 H) = 3.016049 u; M (^4 He) = 4.002603 u, u=931.5 MeV/c^2 )
Q = (M [^3 H] + M [^2 H] − M [^4 He] − M [n])c^2
(c) (5) What is the Fermi-Dirac occupancy factor for an electron 100 mV above the Fermi energy at room temperature? (DATA: kTroom ' 1 /40 eV)
fF D = 1/(e(E−EF^ )/kT^ + 1)
- (20) A (nonrelativistic) particle of mass m and kinetic energy E > 0 comes from the left (x = −∞) and hits a potential barrier of height V 0 and width 2 a centered at the origin:
V (x) =
0 if x < −a (region I) +V 0 if −a < x < a (region II) 0 if x > a (region III)
a) (5) Considering just the case of incoming particles from the left with energy less than V 0 , on the figure below sketch the potential and a wave function for the scattering case, 0 < E < V 0 , in the region − 3 a < x < +3a.
Region II
x/a
Region I
Region III
b) (5) Considering just the 0 < E < V 0 case, write the (time independent) Schroedinger equation which applies in regions I, II, and III.
Region I: − ¯h 2 2 m
d^2 ΨI (x) dx^2 =^ EΨI^ (x)
Region II: − ¯h
2 2 m
d^2 ΨII (x) dx^2 +^ V^0 ΨII^ (x) =^ EΨII^ (x)
Region III:− h¯
2 2 m
d^2 ΨIII (x) dx^2 =^ EΨIII^ (x)
c) (5) Continuing the 0 < E < V 0 case, give the general solution to the Schroedinger equation in regions I, II, and III introducing appropriate constants and defining all parameters to be real.
Region I: ΨI (x) = Aeıkx^ + Be−ıkx, where k =
2 mE ¯h^2
Region II: ΨII (x) = Ceαx^ + De−αx, where α =
2 m(V 0 −E) ¯h^2
Region III: ΨIII (x) = F eıkx^ + Ge−ıkx, where k is the same as that of Region I.
d) (5) Suppose after applying the boundary conditions, you solved for ALL the unknown coefficients in your general solutions above. From these coefficients how would you calculate the probability that the particle tunnels through the barrier?
Since A is the amplitude of the incoming wave and F is the amplitude of the transmitted wave, the probability of transmission through the barrier (tunneling) is: P = |F A |^2.
- (20) (a) (5) Explain the fundamental physics responsible for the band structure of the electronic states in crystals.
The Schroedinger equation determines the allowed states of electrons in metals which share one (or more) electrons to the entire lattice. It turns out that the solutions to the Schroedinger equation for a periodic potential display a band structure.
(b) (5) Explain the electrical conductivity properties of metals and insulators based on band theory.
Given the fact that the allowed states of the electrons in a crystal occur in bands and the fact that electrons obey the Pauli exclusion principle, the electrical conduction properties are determined by the location of the last filled electron (Fermi level). Insulators have large gaps between the valence and conduction bands and the Fermi level falls at the top of the valence band. Thus there are no empty states available nearby for the electrons to accept energy and conduct. For conductors the Fermi level falls within a band so there are empty states available for the electrons to accept energy and conduct. Semiconductors are like insulators only with small band gaps which (through impurities and temperature effects) allows for some electrons to be available for conduction.
(c) (5) Sketch the binding energy per nucleon versus A curve for the STABLE nuclei. Use it to explain how neutron-induced fission of heavy elements such as 235 U can release energy?
A
BE/A
As one can see from the BE/A curve the protons and neutrons in a large nucleus have less binding energy per particle than nuclei in the middle of the curve. Thus splitting a large nucleus into two smaller nuclei will result in a more tightly bound system altoghether and therefore release energy. (d) (5) Sketch the N versus Z curve for the STABLE nuclei. Use it to explain how neutron-induced fission of heavy elements such as 235 U can lead to a chain reaction?
N
Z
As one can see from the N versus Z curve of stable nuclei, the larger nuclei have a larger fraction of neutrons. Thus when a large nucleus fissions, its fission products will have too many neutrons to be stable. The will approach stability by beta-decaying (which turns neutrons into protons) or by releasing neutrons which can then induce more fissions leading to a chain reaction.
