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:

See the attachment

Hope it helps you

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Moment of Inertia $\sf(l)=2\times 10^{-40}\:gm\:cm^{2}$

• Reduced Mass $\sf(\mu)=1.0\:g\:mol^{-1}$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Bond length $\sf,r$

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{ As we know that :}}$

$\dashrightarrow \sf I=\mu r^2$

$\underline{\:\textsf{ Now, put the given values :}}$

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

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

• $\sf Bond \:length, r = 1.41 \times {10}^{ - 20} \: cm$

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