1000 Solved Problems in Modern Physics

(Romina) #1

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 26

2.33 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
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