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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2010 · Paper 2 · Physics · Atomic Physics

Paragraph: The key feature of Bohr's theory of spectrum of hydrogen...

Paragraph: The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.Question: In a CO molecule, the distance between C\mathrm{C} (mass =12amu=12 \mathrm{amu} )andOand \mathrm{O} (mass = 16amu16 \mathrm{amu} ), where 1amu=53×10−27 kg1 \mathrm{amu}=\frac{5}{3} \times 10^{-27} \mathrm{~kg}, is close to

Options

  1. A2.4×10−10 m2.4 \times 10^{-10} \mathrm{~m}
  2. B1.9×10−10 m1.9 \times 10^{-10} \mathrm{~m}
  3. C1.3×10−10 m1.3 \times 10^{-10} \mathrm{~m}
  4. D4.4×10−11 m4.4 \times 10^{-11} \mathrm{~m}

Correct answer

C. 1.3×10−10 m1.3 \times 10^{-10} \mathrm{~m}

Step-by-step solution

I=μr2I=\mu r^2 ( where μ=\mu= reduced mass) μ=m1m2m1+m2=487amu=11.43×10−27 kg\mu=\frac{m_1 m_2}{m_1+m_2}=\frac{48}{7} \mathrm{amu}=11.43 \times 10^{-27} \mathrm{~kg} Substituting in I=μr2I=\mu r^2 we get, r=Iμ=1.87×10−4611.43×10−27=1.28×10−10 m\begin{aligned}r & =\sqrt{\frac{I}{\mu}}=\sqrt{\frac{1.87 \times 10^{-46}}{11.43 \times 10^{-27}}}\\ & =1.28 \times10^{-10} \mathrm{~m} \end{aligned}\therefore$ The correct answer is (c).
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