1000 Solved Problems in Modern Physics

120 2 Quantum Mechanics – I

2.56

n(E 2 )

n(E 1 )

=e−(E^2 −E^1 )/kT=

1

10

T=

E 2 −E 1

kln 10

=

2. 26

8. 625 × 10 −^5 × 2. 3

= 1. 14 × 104 K

2.3.6 Molecules ....................................

2.57 The two modes of motion of a diatomic molecule are (i) rotation and (ii) vibra-
tion.
The first order rotational energy is^2 J(J+1)/ 2 I 0 , whereI 0 =MR^20 is
the moment of inertia of the molecule about an axis perpendicular to the line
joining the nuclei; the energy being the same as for the rigid rotator. Clearly
the spacing between successive levels is unequal; it progressively increases
with the increasing value ofJ, whereJ = 0 , 1 , 2 ...The spectrum called
band spectrum arises due to optical transitions between rotational levels. The
band spectrum is actually a line spectrum, but is thus called because the lines
are so closely spaced and unresolved with an ordinary spectrograph, and give
the appearance of a band.
The second mode consists of to and fro vibrations of the atoms about the
equilibrium position. The motion is described as simple harmonic motion.
The energy levels are given byEn =ω(n+ 1 /2), wheren = 0 , 1 , 2 ...
and are equally spaced. However as J or n increases, the spacing between
levels becomes smaller than that predicted from the simple rigid rotator and
harmonic oscillator.

2.58 The rotational energy levels are given by

EJ=^2 J(J+1)/ 2 Io

whereIois the rotational inertia

ΔE=E 1 −E 0 =^2 /Io

Ifμis the reduced mass,

I=μr^2 =

mHmDr^2

mH+mD

=

mH. 2 mHr^2

mH+ 2 mH

=

(

2

3

)

mHr^2

(becausemD≈ 2 mH)

ΔE=

3 ^2

2 mHr^2

=

3

2

.

(cMeV−fm)^2

mHc^2 (0. 075 × 10 −^9 m)^2

=

3

2

(197. 3 × 10 −^15 MeV−m)^2

938(0. 075 × 10 −^9 m)^2

= 0. 011 × 10 −^6 MeV

= 0 .011 eV.

λ= 1 , 241 / 0. 011 = 1. 128 × 105 nm

= 0 .113 mm

2.59 All nuclei of evenA, with zero or non-zero spin obey Bose statistics and
all those of odd Aobey Fermi statistics. The result has been crucial in