4.3 Solutions
281
4.49 p(E J ) = (2 j + 1)e
−J (J +1) 2
2I 0 kT
The factor
2
2I 0 k
=
(1.055 × 10
−34 )
2
2 × 4.64 × 10 −48 × 1.38 × 10 −23 J
= 86.9
p(E 0 ) = 1
p(E 1 ) = 3e
−2×86.9/400
= 1.942
p(E 2 ) = 5e
−6×86.9/400
= 1.358
p(E 3 ) = 7e
−12×86.9/400
= 0.516
4.50 p(E 2 ) = 5e
−6×86.9/T
= 5e
−521.4/T
(1)
p(E 3 ) = 7e
−12×86.9/T
= 7e
−1042.8/T
Equating p(E 2 ) and p(E 3 ) and solving for T , we find T = 1,549 K
4.51 For Boltzmann statistics p(E) ∝ e
−E/kT Therefore,
p(E n )
p(E 1 )
= e
−(En −E1)/kT
In hydrogen atom, if the ground state energy E 1 = 0, then E 2 = 10.2,
E 3 = 12.09 and E 4 = 12.75 eV
The factor kT = 8.625 × 10
−5
× 6, 000 = 0.5175
P(E 2 )/P(E 1 ) = e
−10.2/0.5175
= 2.75 × 10
−9
P(E 3 )/P(E 1 ) = e
−12.09/0.5175
= 1.4 × 10
−10
P(E 4 )/P(E 1 ) = e
−12.75/0.5175
= 1.99 × 10
−11
Thus P(E 1 ) : P(E 2 ) : P(E 3 ) :: 1 : 2.8 × 10
−9 : 1.4 × 10
−10 : 2.0 × 10
−11
This then means that the hydrogen atoms in the chromospheres are predominantly in the ground state.
4.52 p(E) =
1
e (E−E F )/kT + 1
For E − E F = kT, p(E) =
1
e + 1
= 0.269
For E − E F = 5kT, p(E) =
1
e 5 + 1
= 6.69 × 10
−3
For E − E F = 10kT, p(E) =
1
e 10 + 1
= 4.54 × 10
−5
4.53 For the conduction electrons, the number of states per unit volume with energy
in the range E and E +dE, can be written as n(E)dE where n(E) is the density
of states. Now, for a free electron gas
n(E) =
8
√
2π m
3/2
h 3
E
1/2