108 2 Quantum Mechanics – I
f=me^4 / 4 n^3 h^3 ε 02ν=
me^4
8 ε 02 h^3(
1
n^2 f−
1
n^2 i)
=
me^4
8 ε 02 h^3{
(ni−nf)(ni+nf)
n^2 in^2 f}
If bothniandnfare large, and if we let
ni=nf+ 1 ,ν≈me^4
8 ε^20 h^3(
2
n^3 i)
=f2.20 Energy difference for the transitions in the two seriesΔE 11 −ΔE 32 =
1 , 241 / 16. 58 = 74 .85 eV
13. 6 Z^2{(
1
12
−
1
22
)
−
(
1
22
−
1
32
)}
= 74. 85
Solving forZ, we getZ=3.
The ion isLi++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 betweenn = 4(E 4 =− 0 .85 eV) andn =2(E 2 =− 3 .4eV).
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 betweenn=6(E 6 =− 0 .38 eV) andn=2(E 2 =− 3 .4eV).
ΔE 62 =− 0. 38 −(− 3 .4)= 3 .02 eV
ThusΔE 62 −ΔE 42 = 3. 02 − 2. 55 = 0 .47 eV
The difference of 0.47 eV is also equal to difference inE 6 =−^0 .38 eV
(n=6) andE 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 spec-
troscopy, ̃vmn−v ̃kn=v ̃mk
2.3.3 X-rays ...........................................
2.22 The wavelengthλLK= 0 .0724 nm corresponds to the energy
Eγ= 1 , 241 /λLK= 1 , 241 / 0. 0724 = 17 ,141 eV
Now 17, 141 = 13. 6 ×^34 (Z−σ)^2
The factor 3/4 is due to theL→Ktransition. SubstitutingZ=42, and
solving forσwe obtainσ= 1. 0
2.23 Eγ=^13.^6 ×3(Z−σ)^2 /^4 =^13.^6 ×3(42−1)^2 /^4 =^17 ,^146 .2eV.
λLK= 1 , 241 / 17 , 146. 2 = 0 .07238 nm
= 0. 7238 A ̊2.24 Cobalt:EK= 13 .6(Z−σ)^2 = 13 .6(27−1)^2 = 9 , 193 .6eV
λK=1 , 241
9 , 193. 6
nm= 0 .135 nm= 1. 35 A ̊