Question:

The inter-molecular distance between two atoms of a hydrogen molecule is 0.77A˚0.77 \, \text{\AA}, and the mass of a proton is 1.67×1027Kg1.67 \times 10^{-27} \, \text{Kg}. The moment of inertia of a molecule is:

Show Hint

The moment of inertia for molecular systems is crucial in determining their rotational spectra, which is foundational in molecular physics and chemistry for identifying molecular structures.
Updated On: Mar 17, 2025
  • $4.95 \times 10^{-47} \, \text{Kg·m}^2$
  • $0.495 \times 10^{-47} \, \text{Kg·m}^2$
  • $4.5 \times 10^{-47} \, \text{Kg·m}^2$
  • $45.9 \times 10^{-47} \, \text{Kg·m}^2$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

To calculate the moment of inertia II for a diatomic molecule such as hydrogen (H2H_2), use the formula:
I=μr2I = \mu r^2
where μ\mu is the reduced mass given by μ=m1×m2m1+m2\mu = \frac{m_1 \times m_2}{m_1 + m_2} for two identical masses mm, and rr is the inter-molecular distance. For two protons, each with mass m=1.67×1027kgm = 1.67 \times 10^{-27} \, \text{kg}:
μ=m×m2m=m2=1.67×1027kg2=0.835×1027kg\mu = \frac{m \times m}{2m} = \frac{m}{2} = \frac{1.67 \times 10^{-27} \, \text{kg}}{2} = 0.835 \times 10^{-27} \, \text{kg}
r=0.77×1010m=0.77A˚r = 0.77 \times 10^{-10} \, \text{m} = 0.77 \, \text{\AA}
I=0.835×1027kg×(0.77×1010m)20.495×1047kg-m2I = 0.835 \times 10^{-27} \, \text{kg} \times (0.77 \times 10^{-10} \, \text{m})^2 \approx 0.495 \times 10^{-47} \, \text{kg-m}^2

Was this answer helpful?
0
0