110
2 Quantum Mechanics – I
3 d →
2 P 1/2 ,
3 d 3/2 →
2 P 3/2
3 d 5/2 →
2 P 3/2
In all there are seven allowed transitions.
2.31 Let the wavelength difference be Δλ when voltage V is applied.
1,200
(Z − 1) 2 −
12.4
V
= Δλ
(1)
1,200
(Z − 1) 2 −
12.4
10
= Δλ
(2)
1,200
(Z − 1) 2 −
12.4
20
= 3Δλ
(3)
Note that the first term on the LHS of (1) (Corresponding to the characteristic X-rays) is unaffected due to the application of voltage. Eliminating Δλ
between (2) and (3), we find Z = 28.82 or 29. The target material is Copper.
2.32 λ k =
1,200
(Z − 1) 2
˚
A
Z 1 = 1 +
1,200
2.4
1/2
= 23.36
Z 2 = 1 +
1,200
1.8
1/2
= 26.82
The required elements have Z = 23, 24, 25 and 26
2.33 Bohr’s theory gives
ν =
m e e
4
8ε
2
0 h 3
1
1 2 −
1
n 2
(z − b)
2
(1)
or
√
v =
m e e
4
8ε
2
0 h 3
1
1 2 −
1
n 2
1
2
(z − b)
( 2 )
The factor within the square brackets is identified as a. Substitute m e =
9.11 × 10
−31 kg,
e = 1.6 × 10
−19 C,
ε 0 = 8.85 × 10
−12 F/m and h = 6.626 × 10
−34 J – s and put n = 2 to find a.
We get
a = 4.956 × 10
7 Hz
1/2