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AIEEE 2012 Solved Paper

Section: Physics

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Question : 56 of 89

Marks: +1, -0

A diatomic molecule is made of two masses

m_1

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

$\frac{(m_1+m_2)^2 n^2 h^2}{2 m_1^2 m_2^2 r^2}$
$\frac{n^2 h^2}{2 (m_1+m_2) r^2}$
$\frac{2 n^2 h^2}{(m_1+m_2) r^2}$
$\frac{(m_1+m_2) n^2 h^2}{2 m_1 m_2 r^2}$

Solution:

The energy of the system of two atoms of diatomic

molecule

E=\frac{1}{2} I \omega

where

I=

moment of inertia

\omega=

Angular velocity

=\frac{L}{I}

.

L=

Angular momentum

I=\frac{1}{2} (m_1 r_1^2+m^2 r_2^2)

Thus,

E=\frac{1}{2} (m_1 r_1^2+m_2 r_2^2) \omega^2

......(1)

E=\frac{1}{2} (m_1 r_1^2+m_2 r_2^2) \frac{L^2}{I^2}

L=n \frac{n h}{2 n}

(According Bohr's Hypothesis)

E=\frac{1}{2} (m_1 r_1^2+m_2 r_2^2) \frac{L^2}{(m_1 r_1^2+m_2 r_2)^2)^2}

E=\frac{1}{2} \frac{L^2}{(m_1 r_1^2+m_2 r_2^2)}

=\frac{n^2 h^2}{8 \pi^2 (m_1 r_1^2+m_2 r_2^2)}

E=\frac{(m_1+m_2) n^2 h^2}{8 \pi^2 r^2 m_1 m_2}

[.

as

r_1=\frac{m_2 r}{m_1+m_2} ; r_2=\frac{m_2 r}{m_1+m_2}]

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