The moment of inertia of an HCl molecule abou | vedclass

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(B) The masses of the atoms are $m_1 = 1 \ amu$ (for $H$) and $m_2 = 35.5 \ amu$ (for $Cl$). The interatomic distance is $L = 1 \ \mathring A = 10^{-1

Vedclass · Accepted Answer

$1.61 	imes 10^{-47} \ kg \ m^2$

Vedclass · Answer

(B) The masses of the atoms are $m_1 = 1 \ amu$ (for $H$) and $m_2 = 35.5 \ amu$ (for $Cl$). The interatomic distance is $L = 1 \ \mathring A = 10^{-10} \ m$.Let $x$ be the distance of $H$ from the center of mass ($C$.$M$.). Then the distance of $Cl$ from the $C$.$M$. is $(L - x)$.Using the definition of the center of mass: $m_1 x = m_2 (L - x) \Rightarrow 1 \cdot x = 35.5 \cdot (1 - x)$.$x = 35.5 - 35.5x \Rightarrow 36.5x = 35.5 \Rightarrow x = \frac{35.5}{36.5} \ \mathring A \approx 0.9726 \ \mathring A$.The distance of $Cl$ from the $C$.$M$. is $(L - x) = 1 - 0.9726 = 0.0274 \ \mathring A$.The moment of inertia $I$ is given by $I = m_1 x^2 + m_2 (L - x)^2$.Using $1 \ amu = 1.67 	imes 10^{-27} \ kg$:$I = (1 	imes 1.67 	imes 10^{-27}) 	imes (0.9726 	imes 10^{-10})^2 + (35.5 	imes 1.67 	imes 10^{-27}) 	imes (0.0274 	imes 10^{-10})^2$.$I = 1.67 	imes 10^{-27} 	imes 10^{-20} 	imes [0.9726^2 + 35.5 	imes 0.0274^2]$.$I = 1.67 	imes 10^{-47} 	imes [0.946 + 35.5 	imes 0.00075] \approx 1.67 	imes 10^{-47} 	imes [0.946 + 0.0266] \approx 1.67 	imes 10^{-47} 	imes 0.9726 \approx 1.62 	imes 10^{-47} \ kg \ m^2$.This is approximately $1.61 	imes 10^{-47} \ kg \ m^2$.

The moment of inertia of an $HCl$ molecule about an axis passing through its centre of mass and perpendicular to the line joining the $H^+$ and $Cl^-$ ions is,if the interatomic distance is $1 \ \mathring A$:

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