1000 Solved Problems in Modern Physics

2.3 Solutions 105

(b) Equating the coulomb force to the centripetal force

Ze^2 / 4 πεor^2 =mv^2 /r (2)

Solving (1) and (2)

v=Ze^2 /^2 εonh (3)

r=εon^2 h^2 /πmZe^2 (4)

Total energyE=K+U=

1

2

mv^2 −Ze^2 / 4 πεor (5)

Substituting (3) and (4) in (5)

E=−me^4 Z^2 / 8 ε^2 on^2 h^2 (6)

(c) E=−mZ^2 e^4 / 8 ε^2 on^2 h^2 =− 9 me^4 / 8 ε^2 on^2 h^2 (forZ=3)

(d) Ionization energy forBe+^3 = 13. 6 × 32 = 122. 4 eV

2.12 For a hydrogen-like atom the energy in thenth orbit isEn= 13. 6 μZ^2 /n^2
For hydrogen atom the reduced massμ≈me, while for muon mesic atom
it is of the order of 200me. Consequently, the transition energies are enhanced
by a factor of about 200, so that the emitted radiation falls in the x-ray region
instead of U.V., I.R. or visible part of electromagnetic spectrum. The radius is
given by
rn=ε 0 n^2 h^2 /πμe^2
Here, because of inverse dependence onμ, the corresponding radii are reduced
by a factor of about 200.

2.13r 1 =

a 0

μZ

=R=r 0 A^1 /^3 =r 0 (2Z)^1 /^3

Z^4 /^3 ≈

0. 529 × 10 −^10

207 × 1. 3 × 21 /^3 × 10 −^15

= 156

ThereforeZ= 44 .14 or 44

The first orbit of mu mesic atom will be just grazing the nuclear surface in

the atom of Ruthenium. Actually, in this regionA≈ 2. 2 Zso that the answer

would beZ≈ 43

2.14 The first four lines of the Lyman series are obtained from the transition ener-
gies betweenn= 2 →1, 3→1, 4→1, 5→ 1

NowEn=−

α^2 mec^2

4 n^2