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 beΔλwhen voltageVis 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
= 3 Δλ (3)Note that the first term on the LHS of (1) (Corresponding to the character-
istic X-rays) is unaffected due to the application of voltage. EliminatingΔλ
between (2) and (3), we findZ= 28 .82 or 29. The target material is Copper.2.32λk=
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 haveZ= 23 , 24 ,25 and 262.33 Bohr’s theory gives
ν=mee^4
8 ε^20 h^3(
1
12
−
1
n^2)
(z−b)^2 (1)or√
v=[
mee^4
8 ε^20 h^3(
1
12
−
1
n^2)]^12
(z−b)(2)The factor within the square brackets is identified as a. Substituteme =
9. 11 × 10 −^31 kg,
e= 1. 6 × 10 −^19 C,
ε 0 = 8. 85 × 10 −^12 F/m andh= 6. 626 × 10 −^34 J – s and putn=2 to finda.
We get
a= 4. 956 × 107 Hz^1 /^2