A diatomic molecule is made of two masses m_1 | vedclass

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Question

(A) The rotational kinetic energy $(RKE)$ of a system of two masses $m_1$ and $m_2$ separated by distance $r$ rotating about their center of mass is g

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The rotational kinetic energy $(RKE)$ of a system of two masses $m_1$ and $m_2$ separated by distance $r$ rotating about their center of mass is given by $RKE = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.The reduced mass is $\mu = \frac{m_1 m_2}{m_1 + m_2}$.The moment of inertia about the center of mass is $I = \mu r^2 = \frac{m_1 m_2}{m_1 + m_2} r^2$.According to Bohr's quantization rule,the angular momentum is $L = \frac{nh}{2\pi}$.Substituting these into the $RKE$ formula:$RKE = \frac{(nh/2\pi)^2}{2I} = \frac{n^2 h^2}{4\pi^2 \cdot 2 \cdot (\frac{m_1 m_2}{m_1 + m_2} r^2)} = \frac{n^2 h^2 (m_1 + m_2)}{8\pi^2 m_1 m_2 r^2}$.Note: The provided options in the prompt appear to have typographical errors regarding the $\pi$ factor. Based on standard physics,the correct expression is $\frac{(m_1 + m_2) n^2 h^2}{8 \pi^2 m_1 m_2 r^2}$.

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:

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