1000 Solved Problems in Modern Physics

4.3 Solutions 281

4.49 p(EJ)=(2j+1)e

−J(J+1)^2

2 I 0 kT

The factor

^2

2 I 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 )= 3 e−^2 ×^86.^9 /^400 = 1. 942

p(E 2 )= 5 e−^6 ×^86.^9 /^400 = 1. 358

p(E 3 )= 7 e−^12 ×^86.^9 /^400 = 0. 516

4.50 p(E 2 )= 5 e−^6 ×^86.^9 /T= 5 e−^521.^4 /T (1)

p(E 3 )= 7 e−^12 ×^86.^9 /T= 7 e−^1042.^8 /T

Equatingp(E 2 ) andp(E 3 ) and solving forT, we findT= 1 ,549 K

4.51 For Boltzmann statisticsp(E)∝e−E/kTTherefore,

p(En)

p(E 1 )

=e−(En−E^1 )/kT

In hydrogen atom, if the ground state energyE 1 = 0 ,thenE 2 = 10. 2 ,

E 3 = 12 .09 andE 4 = 12 .75 eV

The factorkT= 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

ThusP(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 predomi-

nantly in the ground state.

4.52 p(E)=

1

e(E−EF)/kT+ 1

ForE−EF=kT,p(E)=

1

e+ 1

= 0. 269

ForE−EF= 5 kT,p(E)=

1

e^5 + 1

= 6. 69 × 10 −^3

ForE−EF= 10 kT,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 rangeEandE+dE, can be written asn(E)dEwheren(E) is the density
of states. Now, for a free electron gas

n(E)=

8

√

2 πm^3 /^2

h^3

E^1 /^2