80. Letting
T
≈ e
−2kL
= exp
−2L
8π
2
m(U
− E)
h
2
,
and using the result of Exercise 3 in Chapter 39, we solve for E:
E
=
U
−
1
2m
h ln T
4πL
2
=
6.0 eV
−
1
2(0.511 MeV)
(1240 eV
·nm)(ln 0.001)
4π(0.70 nm)
2
=
5.1 eV .