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Physics / An Introduction to Thermal Physics 1 / Chapter A / Problem 15P

An Introduction to Thermal Physics | 1st Edition | ISBN: 9780201380279 | Authors: Daniel V. Schroeder

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A CO molecule can vibrate with a natural frequency of 6.4

Chapter , Problem 15P

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QUESTION:

Problem 15P

A CO molecule can vibrate with a natural frequency of 6.4 × 1013 s−1.

(a) What aret energies (in eV) of the five lowest vibrational slates of a CO molecule?

(b) If a CO molecule is initially in its ground stale and you wish to excite it into its first vibrational level, what wavelength of light should you aim at it?

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QUESTION:

Problem 15P

A CO molecule can vibrate with a natural frequency of 6.4 × 1013 s−1.

(a) What aret energies (in eV) of the five lowest vibrational slates of a CO molecule?

(b) If a CO molecule is initially in its ground stale and you wish to excite it into its first vibrational level, what wavelength of light should you aim at it?

ANSWER:

Solution 15P

A CO molecule is vibrating with a natural frequency of 6.4 × ${10}^{13}$ ${s}^{-1}$ .

The vibrational energy for a CO molecule in a $n\$ state is given by

${E}_{n}$ = $(n+1/2)\ h$ $\nu{}$

Where, $\nu{}$ = frequency of vibration

$h$ = Planck’s constant = 6.62 $\times{}{10}^{-34}$ J/s

$n=0,1,2,3$ …

Step 1 of 1

The energies (in eV) of the five lowest vibrational states of a CO molecule can be found by replacing n with 0,1,2,3 ,4 and substituting for $h$ and $\nu{}$ .

${E}_{0}$ = $(0+1/2)\ h$ $\nu{}$ = $\frac{1}{2}$ $h\nu{}$ = $\frac{1}{2}$ $(6.62\times{}{10}^{-34})(6.4\ ×\ {10}^{13})$ =2.12 $\times{}{10}^{20}$ J =0.1323 eV

${E}_{1}$ = $(1+1/2)\ h$ $\nu{}$ = $\frac{3}{2}$ $h\nu{}$ = $3\times{}$ $\frac{1}{2}$ $h\nu{}$ = 3 $\times{}$ 0.1323= 0.397 eV

${E}_{2}$ = $(2+1/2)\ h$ $\nu{}$ = $\frac{5}{2}$ $h\nu{}$ = $5\times{}$ $\frac{1}{2}$ $h\nu{}$ = 5 $\times{}0.1323$ = 0.6617 eV

${E}_{3}$ = $(3+1/2)\ h$ $\nu{}$ = $\frac{7}{2}$ $h\nu{}$ = $7\times{}$ $\frac{1}{2}$ $h\nu{}$ =7 $\times{}$ 0.1323= 0.926 eV

${E}_{4}$ = $(4+1/2)\ h$ $\nu{}$ = $\frac{9}{2}$ $h\nu{}$ = $9\times{}$ $\frac{1}{2}$ $h\nu{}$ = 9 $\times{}$ 0.1323= 1.19 eV

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