# A diatomic molecule is made of two masses  and which are separated by a distance . If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by ( is an integer) Option 1) Option 2) Option 3) Option 4)

A diatomic molecule consists of two atoms of  masses $m_{1}$ and $m_{2}$ at a distance $r$ apart. Let $r_{1}$and $r_{2}$  be the distances of the atoms from the centre of mass

The moment of inertia of this molecule about an axis passing through its centre of mass and perpendicular to a line joining the atoms is

The solution is correct. So no need to change it

$I= m_{1}r_{1}^{2}+m_{2}r_{2}^{2}$

$As\: \: \: m_{1}r_{1}= m_{2}r_{2}\: \: \:or\: \: r_{1}= \frac{m_{2}}{m_{1}}r_{2}$

$\because \: \: r_{1}+r_{2}= r$

$\therefore \: \: r_{1}= \frac{m_{2}}{m_{1}}\left ( r-r_{1} \right )$

On rearranging, we get

$r_{1}= \frac{m_{2}r}{m_{1}+m_{2}}$

$Similarly \: \: r_{2}= \frac{m_{1}r}{m_{1}+m_{2}}$

$Therefore,\: the \: moment\: of\: inertia\: can \: be\: written\: as$

$I= m_{1}\left ( \frac{m_{2}r}{m_{1}+m_{2}} \right )^{2}+m_{2}\left ( \frac{m_{1}r}{m_{1}+m_{2}} \right )^{2}$

$= \frac{m_{1}m_{2}}{m_{1}+m_{2}}r^{2}\cdots \cdots \cdots \cdots (i)$

According to Bohr’s quantisation condition

$L= \frac{nh}{2\pi }$

$or\: \: L^{2}= \frac{n^{2}h^{2}}{4\pi ^{2}}$

$Rotational\: \: energy, E= \frac{L^{2}}{2l}$

$E= \frac{n^{2}h^{2}}{8\pi ^{2}I}\cdots \cdots \cdots \cdots using(ii)$

$E= \frac{n^{2}h^{2}\left ( m_{1}+m_{2} \right )}{8\pi ^{2}\left ( m_{1}m_{2} \right )r^{2}}\cdots \cdots \cdots \cdots using(i)$

$= \frac{n^{2}h^{2}\left ( m_{1}+m_{2} \right )}{2m_{1}m_{2} r^{2}}$

In the question instead of  h, should be given.

Option 1)

Incorrect

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Correct

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