2.3 Solutions 109
Iron:EK= 13 .6(26−1)^2 = 8 ,500 eV
λK=1 , 241
8 , 500
= 0 .146 nm= 1. 46 A ̊2.25 The minimum wavelength of the photon will correspond to maximum fre-
quency which will be determined byE=hvmax
λmin=c
vmax=
hc
hvmax=
hc
E=
2 πc
E=2 π× 197 .3fm−MeV
30 × 10 −^3 MeV= 4. 13 × 105 fm= 4. 13 A ̊2.26λC=
hc
eV
h=eVλC
c=
1. 6 × 10 −^19 × 80 × 103 × 0. 15 × 10 −^10
3 × 108
= 6. 4 × 10 −^34 J−s2.27λc=
hc
eV
h=λceV
c=
0. 247 × 10 −^10 × 1. 6 × 10 −^19 × 50 , 000
3 × 108
= 6. 59 × 10 −^34 J−s2.28 According to Mosley’s law
1
λ
=A(Z−1)^2
1
λI=A(26−1)^2
1
λCu=A(29−1)^2
λCu
λI=
252
282
= 0. 797 →λCu= 193 × 0. 797 = 153 .8pm2.29 λK−λC=84 pm= 0. 84 A(1) ̊
1 , 200
(28−1)^2−
12. 4
V
= 0. 84 (2)
whereλC=hc
eV=
12. 4
V
(V is in kV) (3)Solving forVin (2),V= 15 .4kV2.30 TheLαline is produced due to transitionn= 3 →n=2. For then=2 shell
the quantum numbers arel=0orl=1 andj=l±^12 , the energy states
being^2 S 1 / 2 ,^2 P 1 / 2 ,^2 P 3 / 2 .Forn=3 shell the energy states are^3 S 1 / 2 ,^3 P 1 / 2 ,
(^3) P 3 / 2 , (^3) d 3 / 2 , (^3) d 5 / 2
The allowed transitions are
(^3) S
1 / 2 →
(^2) P
1 / 2 ,
(^3) S
1 / 2 →
(^2) P
3 / 2
(^3) P
1 / 2 →
(^2) S
1 / 2 ,
(^3) P
3 / 2 →
(^2) S
1 / 2