2.3 Solutions
109
Iron: E K = 13.6(26 − 1)
2
= 8,500 eV
λ K =
1,241
8,500
= 0.146 nm = 1.46 ˚
A
2.25 The minimum wavelength of the photon will correspond to maximum frequency which will be determined by E = hv max
λ min =
c
v max
=
hc
hv max
=
hc
E
=
2πc
E
=
2π × 197.3 fm − MeV
30 × 10 −3 MeV
= 4.13 × 10
5 fm = 4.13 ˚
A
2.26 λ C =
hc
eV
h =
eV λ C
c
=
1.6 × 10
−19
× 80 × 10
3
× 0.15 × 10
−10
3 × 10 8
= 6.4 × 10
−34 J − s
2.27 λ c =
hc
eV
h =
λ c eV
c
=
0.247 × 10
−10
× 1.6 × 10
−19
× 50,000
3 × 10 8
= 6.59 × 10
−34 J − s
2.28 According to Mosley’s law
1
λ
= A(Z − 1)
2
1
λ I
= A(26 − 1)
2
1
λ Cu
= A(29 − 1)
2
λ Cu
λ I
=
25
2
28 2 = 0.797 → λ Cu = 193 × 0.797 = 153.8 pm
2.29 λ K − λ C = 84 pm = 0.84 ˚
A
( 1 )
1,200
(28 − 1) 2 −
12.4
V
= 0.84
(2)
where λ C =
hc
eV
=
12.4
V
(V is in kV)
(3)
Solving for V in (2), V = 15.4 kV
2.30 The L α line is produced due to transition n = 3 → n = 2. For the n = 2 shell
the quantum numbers are l = 0 or l = 1 and j = l ±
1
2
, the energy states
being
2 S 1/2 ,
2 P 1/2 ,
2 P 3/2 . For n = 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