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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Physics
Chapter
Atomic Physics

JEE Main 2012 · Offline · Physics · Atomic Physics

A diatomic molecule is made of two masses m_1 and m_2 which are...

A diatomic molecule is made of two masses m1m_1 and m2m_2 which are separated by a distance rr. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by ( nn is an integer)

Options

  1. A(m1+m2)2n2h22m12m22r2\frac{\left(m_1+m_2\right)^2 n^2 h^2}{2 m_1^2 m_2^2 r^2}
  2. Bn2h22(m1+m2)r2\frac{n^2 h^2}{2\left(m_1+m_2\right) r^2}
  3. C2n2h2(m1+m2)r2\frac{2 n^2 h^2}{\left(m_1+m_2\right) r^2}
  4. D(m1+m2)n2h22m1m2r2\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}

Correct answer

D. (m1+m2)n2h22m1m2r2\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}

Step-by-step solution

r1=m2rm1+m2;r2=m1rm1+m2(l1+l2)ω=nh2π=nhK.E=12(l1+l2)ω2=n2h2(m1+m2)2m1m2r2r_1=\frac{m_2 r}{m_1+m_2} ; r_2=\frac{m_1 r}{m_1+m_2} \left(l_1+l_2\right) \omega=\frac{n h}{2 \pi}=nh K . E=\frac{1}{2}\left(l_1+l_2\right) \omega^2=\frac{n^2 h^2\left(m_1+m_2\right)}{2 m_1 m_2 r^2}
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