Phys 132 Homework 12 Solutions
Professor Crystal Martin
TA: Ellie Hadjiyska
Problem 12: (Phillips Problem 4.1)
We want to find the classical distance of closest approach of two protons with an energy of
approach equal to 2 keV. (Note: Since the rest mass of a proton is about 938 MeV, we are in the
classical regime and can use the fact that the total energy Eis equal to the Coulomb potential
energy P E at closest approach) or
E= 2 keV = e2
4π0r
or
r=1
4π0
e2
E= (8.9876 ×109N m2C−2)(1.6×10−19 C)2
2000 eV
1 eV
1.6×10−19 J= 7.2×10−13 m
or 0.0072˚
A (compare to ∼1˚
A size of atom) or 720 fm (compare to the ∼1 fm size of proton). The
probability that the two protons penetrate the barrier keeping them apart is dependent on the
Gamov energy (Phillips Eq. (4.10)) given by
EG= (παZAZB)22mrc2
where αis the fine structure constant given by
α=e2
4π0~c
≈1
137
For two protons, ZA=ZB= 1 and the reduced mass is given by
mr=mAmB
mA+mB
=mp
2
The Gamov energy is then
EG= (πα)2mpc2= 493 keV
The probability Pthat the two protons penetrate the Coulomb barrier (Phillips Eq. (4.12)) is
therefore
P≈exp "−EG
E1/2#= exp[−493 keV
2 keV 1/2
] = 1.5×10−7
For two 4He nuclei, ZA=ZB= 2 and the reduced mass is given by
mr=mAmB
mA+mB
=mHe
2≈2mp
