1000 Solved Problems in Modern Physics

106 2 Quantum Mechanics – I

ΔEn, 1 =

α^2 mec^2

4

(

1

12

−

1

n^2

)

,n= 2 , 3 , 4 , 5

Thus,

ΔE 21 =

(

1

137

) 2 (

0. 511 × 106

4

)(

1 −

1

4

)

= 5 .1eV

λ 21 =

1 , 241

5. 1

= 243 .3nm= 2 , 433 A ̊

The wavelengths of the other three lines can be similarly computed. They are

2,053, 1,946 and 1,901A. ̊

2.15 (a) Using Bohr’s theory of hydrogen atom

En=−

μe^4

8 ε^20 h^2 n^2

(1)

whereμis the reduced mass. But the fine structure constant

α=

e^2

4 πε 0 c

(2)

Combining (1) and (2)

En=−

α^2 μc^2

2 n^2

(3)

For positronium,=me/2. Therefore for positron

En=−

α^2 mec^2

4 n^2

(4)

(b) rn=ε^0

n^2 h^2

πμc^2

(5)

rn∝

1

μ

=

2

me

Therefore the radii are doubled.

(c)En∝μ=

me

2

Therefore the transition energies are halved.

2.16 (a)U(r)=−

∫

f(r)dr=

∫

krdr+C

=^12 kr^2 +C

U(0)= 0 →C= 0

U(r)=^12 kr^2

(b) Bohr’s assumption of quantization of angular momentum gives

mvr=n (1)

Equating the attracting force to the centripetal force.