108
2 Quantum Mechanics – I
f = me
4
/4n
3 h
3
ε
2
0
ν =
me
4
8ε
2
0 h 3
1
n
2
f
−
1
n
2
i
=
me
4
8ε
2
0 h 3
(n i − n f )(n i + n f )
n
2
i n
2
f
If both n i and n f are large, and if we let
n i = n f + 1, ν ≈
me
4
8ε
2
0 h 3
2
n
3
i
= f
2.20 Energy difference for the transitions in the two series ΔE 11 − ΔE 32 =
1,241/16.58 = 74.85 eV
13.6Z
2
1
1 2 −
1
2 2
−
1
2 2 −
1
3 2
= 74.85
Solving for Z , we get Z = 3.
The ion is Li
++
2.21 Note that wave number is proportional to energy. The wavelength 486.1 nm
in the Balmer series to the energy difference of 2.55 eV, and is due to the
transition between n = 4(E 4 = −0.85 eV) and n = 2(E 2 = −3.4 eV).
ΔE 42 = −0.85 − (−3.4) = 2.55 eV. The wavelength 410.2 nm in the
Balmer series corresponds to the energy difference of 3.0 eV and is due to
the transition between n = 6(E 6 = −0.38 eV) and n = 2(E 2 = −3.4 eV).
ΔE 62 = −0.38 − (−3.4) = 3.02 eV
Thus ΔE 62 − ΔE 42 = 3.02 − 2.55 = 0.47 eV
The difference of 0.47 eV is also equal to difference in E 6 = −0.38 eV
(n = 6) and E 4 = −0.85 eV (n = 4). Thus the line arising from the transition
n = 6 → n = 4, must belong to Bracket series.
Note that in the above analysis we have used the well known law of spectroscopy, ˜
v mn − ˜
v kn = ˜
v mk
2.3.3 X-rays
2.22 The wavelength λ L K = 0.0724 nm corresponds to the energy
E γ = 1,241/λ L K = 1,241/0.0724 = 17,141 eV
Now 17,141 = 13.6 ×
3
4
(Z − σ )
2
The factor 3/4 is due to the L → K transition. Substituting Z = 42, and
solving for σ we obtain σ = 1.0
2.23 E γ = 13.6 × 3(Z − σ )
2
/4 = 13.6 × 3(42 − 1)
2
/4 = 17,146.2 eV.
λ L K = 1,241/17,146.2 = 0.07238 nm
= 0.7238 ˚
A
2.24 Cobalt: E K = 13.6(Z − σ )
2
= 13.6(27 − 1)
2
= 9,193.6 eV
λ K =
1,241
9,193.6
nm = 0.135 nm = 1.35 ˚
A