# Quantum Chemistry

Chapter10 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44 - 45

Exercise. This is the solution to exercise 10.1 in the book.

Solution. From

${E}_{J}=\frac{{\hslash }^{2}}{2I}J\left(J+1\right)$

we get

$\Delta E=\frac{{\hslash }^{2}}{I}\left(J+1\right)$

or, replacing $\Delta E$ we get

$h\nu =\frac{{h}^{2}}{4{\pi }^{2}I}\left(J+1\right)$

or for $J=0$

$I=\frac{h}{4{\pi }^{2}\nu }$

Replacing with numerical values we get:

$I=2.643\cdot {10}^{-47}\phantom{\rule{1em}{0ex}}{kg\phantom{\rule{1em}{0ex}}m}^{2}$

For this molecules

$\mu =\frac{1\cdot 35}{1+35}\cdot 1.67265\cdot {10}^{-27}\phantom{\rule{1em}{0ex}}kg=1.6262\cdot {10}^{-27}\phantom{\rule{1em}{0ex}}kg$

Since

$I=\mu {R}_{0}^{2}$

then

${R}_{0}=1.275Å$