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  • The wave numbers of the lines in a band are given by v=1000(2 n-1) for n positive and by v=-1000(2 n+1) for  n negative. The moment of inertia of the emitter molecule. of the spectrum.</p

The wave numbers of the lines in a band are given by v=1000(2 n-1) for n positive and by v=-1000(2 n+1) for  n negative. The moment of inertia of the emitter molecule. of the spectrum.

Option: 1

2.8 \times 10^{-49} \mathrm{~kg} \mathrm{~m}^2


Option: 2

2.8 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2


Option: 3

2.8 \times 10^{-48} \mathrm{~kg} \mathrm{~m}^2


Option: 4

2.8 \times 10^{-40} \mathrm{~kg}_{-\mathrm{m}^{-1}}


Answers (1)

best_answer

\mathrm{ \begin{aligned} v & =1000(2 n-1) \\ & =1000,3000,5000, \ldots . . \mathrm{cm}^{-1}, \text { for } n=1,2,3, \ldots \\ v & -1000(2 n+1) \\ & =1000,3000,5000, \ldots . \mathrm{cm}^{-1}, \text { for } n=-1,-2,-3, \ldots \end{aligned} }
This means that the separation between any tin consecutive lines is 2000 \mathrm{ \mathrm{~cm}^{-1} }so that
\mathrm{ \begin{aligned} & 2 B=2000 \mathrm{~cm}^{-1} \\ & B=1000 \mathrm{~cm}^{-1}=10^5 \mathrm{~m}^{-1} \end{aligned} }
The moment of inertia of the molecule is therefore
\mathrm{ \begin{aligned} I=\frac{h}{3 x^2 A C} & =\frac{6.63 \times 10^{-14}}{A \times(3.14)^2 \times\left(10^5\right) \times 3 \times 1.0} \\ & =2.8 \times 10^{-49} \mathrm{kg} m^{2} \end{aligned} }

 

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Ritika Harsh

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