Question

Calculate the bond length of HCL molecules of its moment of inertia 2×10-40 gm cm2 and reduced mass 1.0 g mol-1

Answer 1

Answer:

$\boxed{\mathfrak{Bond \ length \ of \ HCl \ molecule \ (r) = 1.41 \times 10^{-20}}}$

Explanation:

Moment of inertia of a diatomic molecule about an axis passing through the center of the diatomic molecule and perpendicular to bond length is product of reduced mass of molecule and square of bond length i.e.

$\boxed{ \bold{I = \mu r^2}}$

r → Bond length

According to the question,

Moment of inertia (I) = $\sf 2 \times 10^{-40}$ g.cm²

Reduced mass ($\sf \mu$) = 1.0 g/mol

By substituting values in the formula we get:

$\rm \implies 2 \times {10}^{ - 40} = 1 \times {r}^{2} \\ \\ \rm \implies r = \sqrt{2 \times {10}^{ - 40} } \\ \\ \rm \implies r = 1.41 \times {10}^{ - 20} \: cm$

Answer 2

Answer:

Moment of Inertia is product of reduced mass of molecule and square of bond length.

• Moment of Inertia $\sf(I)=2\times 10^{-40}\:gm\:cm^{2}$
• Reduced Mass $\sf(\mu)=1.0\:g\:mol^{-1}$
• Bond length $\sf(r)=?$

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf I=\mu r^2\\\\\\:\implies\sf2 \times {10}^{ - 40} = 1 \times {r}^{2}\\\\\\:\implies\sf r = \sqrt{2 \times {10}^{ - 40} } \\\\\\:\implies\sf r = 1.41 \times {10}^{ - 20} \: cm$