SATHEE: Atoms And Nuclei Question 308

Atoms And Nuclei Question 308

Question: 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)

Options:

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

Show Answer

Answer:

Correct Answer: D

Solution:

The energy of the system of two atoms of diatomic molecule $E = \frac{1}{2} I ω^{2}$ where I=moment of inertia $ω =$ Angular velocity $= \frac{L}{I},$ L=Angular momentum $I = \frac{1}{2} (m_{1} r_{1}^{2} + m_{2} r_{2}^{2})$ Thus, $E = \frac{1}{2} (m_{1} r_{1}^{2} + m_{2} r_{2}^{2}) ω^{2}$ ?.(i) $E = \frac{1}{2} (m_{1} r_{1}^{2} + m_{2} r_{2}^{2}) \frac{L^{2}}{I^{2}}$ ?(ii) $L = n, ℏ$ (According Bohr’s Hypothesis) $E = \frac{1}{2} (m_{1} r_{1}^{2} + m_{2} r_{2}^{2}) \frac{L^{2}}{{(m_{1} r_{1}^{2} + m_{2} r_{2}^{2})}^{2}}$

$E = \frac{1}{2} \frac{L^{2}}{(m_{1} r_{1}^{2} + m_{2} r_{2}^{2})} = \frac{n^{2} h^{2}}{8 π^{2} r^{2} m_{1} m_{2}}$

$E = \frac{1}{2} \frac{L^{2}}{(m_{1} r_{1}^{2} + m_{2} r_{2}^{2})} = \frac{n^{2} h^{2}}{8 π^{2} r^{2} m_{1} m_{2}}$

$E = \frac{(m_{1} + m_{2}) n^{2} h^{2}}{8 π^{2} r^{2} m_{1} m_{2}}$

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