2.3 Solutions
105
(b) Equating the coulomb force to the centripetal force
Ze
2
/4πε o r
2
= mv
2
/r
(2)
Solving (1) and (2)
v = Ze
2
/2ε o nh
(3)
r = ε o n
2 h
2
/π m Ze
2
(4)
Total energyE = K + U =
1
2
mv
2
− Ze
2
/4πε o r
(5)
Substituting (3) and (4) in (5)
E = −me
4 Z
2
/8ε
2
o n
2 h
2
(6)
(c) E = −m Z
2 e
4
/8ε
2
o n
2 h
2
= −9me
4
/8ε
2
o n
2 h
2 (for Z = 3)
(d) Ionization energy for Be
+3
= 13.6 × 3
2
= 122.4 eV
2.12 For a hydrogen-like atom the energy in the nth orbit is E n = 13.6 μZ
2
/n
2
For hydrogen atom the reduced mass μ ≈ m e , while for muon mesic atom
it is of the order of 200 m e . 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
r n = ε 0 n
2 h
2
/πμe
2
Here, because of inverse dependence on μ, the corresponding radii are reduced
by a factor of about 200.
2.13 r 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 × 2 1/3 × 10 −15 = 156
Therefore Z = 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 region A ≈ 2.2 Z so that the answer
would be Z ≈ 43
2.14 The first four lines of the Lyman series are obtained from the transition energies between n = 2 → 1, 3 → 1, 4 → 1, 5 → 1
Now E n = −
α
2 m e c
2
4n 2
105
(b) Equating the coulomb force to the centripetal force
Ze
2
/4πε o r
2
= mv
2
/r
(2)
Solving (1) and (2)
v = Ze
2
/2ε o nh
(3)
r = ε o n
2 h
2
/π m Ze
2
(4)
Total energyE = K + U =
1
2
mv
2
− Ze
2
/4πε o r
(5)
Substituting (3) and (4) in (5)
E = −me
4 Z
2
/8ε
2
o n
2 h
2
(6)
(c) E = −m Z
2 e
4
/8ε
2
o n
2 h
2
= −9me
4
/8ε
2
o n
2 h
2 (for Z = 3)
(d) Ionization energy for Be
+3
= 13.6 × 3
2
= 122.4 eV
2.12 For a hydrogen-like atom the energy in the nth orbit is E n = 13.6 μZ
2
/n
2
For hydrogen atom the reduced mass μ ≈ m e , while for muon mesic atom
it is of the order of 200 m e . 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
r n = ε 0 n
2 h
2
/πμe
2
Here, because of inverse dependence on μ, the corresponding radii are reduced
by a factor of about 200.
2.13 r 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 × 2 1/3 × 10 −15 = 156
Therefore Z = 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 region A ≈ 2.2 Z so that the answer
would be Z ≈ 43
2.14 The first four lines of the Lyman series are obtained from the transition energies between n = 2 → 1, 3 → 1, 4 → 1, 5 → 1
Now E n = −
α
2 m e c
2
4n 2