JEE Main · 2012 · 29 Apr 2012 · Offline · Modern Physics

A diatomic molecule is made of two masses

m_{1}

and

m_{2}

which are separated by a distance

r

. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by ( n is an integer)

(A) $\frac{( m _{1} + m _{2} ) ^{2} n ^{2} h ^{2}}{2 m _{1}^{2} m _{2}^{2} r ^{2}}$
(B) $\frac{n ^{2} h ^{2}}{2 ( m _{1} + m _{2} ) r ^{2}}$
(C) $\frac{2 n ^{2} h ^{2}}{( m _{1} + m _{2} ) r ^{2}}$
(D) $\frac{( m _{1} + m _{2} ) n ^{2} h ^{2}}{2 m _{1} m _{2} r ^{2}}$ ✓ Correct

Answer: (D)

Solution

Treat the molecule as two masses turning about their common centre of mass; the moment of inertia reduces to

I = μ r^{2}

with reduced mass

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

. Bohr's rule quantises the angular momentum as

L = I ω = n ℏ

, and the rotational energy is

E = \frac{L ^{2}}{2 I} = \frac{n ^{2} ℏ ^{2}}{2 I}

. Substituting

I = \frac{m _{1} m _{2} r ^{2}}{m _{1} + m _{2}}

gives

E = \frac{( m _{1} + m _{2} ) n ^{2} ℏ ^{2}}{2 m _{1} m _{2} r ^{2}}

, which in the notation of the choices (writing

h

for the reduced constant) is

\frac{( m _{1} + m _{2} ) n ^{2} h ^{2}}{2 m _{1} m _{2} r ^{2}}

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option