Question

The moment of inertia of HCl molecule about an axis passing through its centre of mass and perpendicular to the line joining the $H^{+}$ and $Cl^{-}$ions will be, if the interatomic distance is 1 Å

• A. $0.61 \times 10^{- 47}kg.m^{2}$
• B. $1.61 \times 10^{- 47}kg.m^{2}$
• C. $0.061 \times 10^{- 47}kg.m^{2}$
• D. 0

Answer 1

Answer: (B)

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

Answer 2

If $r_{1}$ and $r_{2}$ are the respective distances of particles $m_{1}$ and $m_{2}$ from the centre of mass then

$m_{1}r_{1} = m_{2}r_{2}$ ⇒ $1 \times x = 35.5 \times (L - x)$ ⇒ $x = 35.5(1 - x)$

⇒ $x = 0.973Å$ and $L - x = 0.027Å$

Moment of inertia of the system about centre of mass

$I = m_{1}x^{2} + m_{2}(L - x)^{2}$

$I = 1amu \times (0.973Å)^{2} + 35.5amu \times (0.027Å)^{2}$Substituting 1 a.m.u. = 1.67 × 10^–27 kg and 1 Å

= 10^–10 mI $= 1.62 \times 10^{- 47}kgm^{2}$