16.
(a) From Fig. 41-10 and Eq. 41-18,
∆E = 2µ
B
B =
2(9.27
× 10
−24
J/T)(0.50 T)
1.60
× 10
−19
J/eV
= 58 µeV .
(b) From ∆E = hf we get
f =
∆E
h
=
9.27
× 10
−24
J
6.63
× 10
−34
J
·s
= 1.4
× 10
10
Hz = 14 GHz .
(c) The wavelength is
λ =
c
f
=
2.998
× 10
8
m/s
1.4
× 10
10
Hz
= 2.1 cm ,
which is in the short radio wave region.