1000 Solved Problems in Modern Physics

(Grace) #1

2.3 Solutions 115


The magnetic momentμ 0 produced is equal to this current multiplied by
the area enclosed.

μB=iA=ωe.

πr^2
2 π

(2)

Using (1) in (2)

μB=

e
2 me

(3)

μBis known as Bohr magneton.
Electron with total angular momentum [j(j+1)]^1 /^2 has a magnetic
momentμ=[j(j+1)]^1 /^2 μB
The z-component of the magnetic moment isμJ=mJμB
wheremJis thez-component of the angular momentum.

2.3.5 Spectroscopy ....................................


2.45 The principle quantum numberndenotes the number of stationary states in
Bohr’s atom model.n= 1 , 2 , 3 ...
lis called azimuthal or orbital angular quantum number. For a given value of
n,ltakes the values 0, 1 , 2 ...n− 1
The quantum numberml, called the magnetic quantum number, takes the val-
ues−l,−l+ 1 ,−l+ 2 ,...,+lfor a given pair of n and / values. This gives
the following scheme:


n 1 2 3
l 0 010 1 2
ml 0 0 − 10 + 1 0 − 10 + 1 − 2 − 10 + 1 + 2
n 4
l 0 1 2 3
ml 0 − 10 + 1 − 2 − 10 + 1 + 2 − 3 − 2 − 10 + 1 + 2 + 3

ms, the projection of electron spin along a specified axis can take on two values
± 1 /2.
Hence the total degeneracy is 2n^2 .Forn= 3 , 2 × 32 = 18 electrons can be
accommodated.

2.46 According to Laporte rule, transitions via dipole radiation are forbidden
between atomic states with the same parity. This is because dipole moment
has odd parity and the integral


∫∞

−∞ψ


f (dipole moment)ψidτ will vanish
between symmetric limits because the integrand will be odd whenψiandψf
have the same parity. Now the parity of the state is determined by the factor
(−1)l. Thus for the given terms thel-values and the parity are as below
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