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 r - Physics MCQ AIEEE 2012 2026 | ExamTest.live

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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):

Solution & Step-by-step Explanation

The moment of inertia of a diatomic molecule about its center of mass is I = μ r^{2}, where μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} is the reduced mass.Rotational energy E = \frac{L ^{2}}{2 I} .According to Bohr's quantization rule, L = n ℏ . E = \frac{( n ℏ ) ^{2}}{2 μ r ^{2}} = \frac{n ^{2} ℏ ^{2}}{2 ( \frac{m _{1} m _{2}}{m _{1} + m _{2}} ) r ^{2}} = \frac{( m _{1} + m _{2} ) n ^{2} ℏ ^{2}}{2 m _{1} m _{2} r ^{2}} Note: ℏ = \frac{h}{2 π} .

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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):

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

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

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

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

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