Question

Use the data from Figure 40.24 to calculate the first three vibrational energy levels of a $C=O$carbon-oxygen double bond.

Short answer

Energy of $n=1$state = $0.107eV$

Energy of $n=2state=0.321eV$

Energy of$n=3state=0.535eV$

Step-by-step solution

Step 1. Given information

Energy level of harmonic quantum oscillator,

$E_n=\left(n-\frac{1}{2}\right)\frac{h\omega }{2\pi }$

Here,

n denotes the state,

h = Planck's constant and

$\omega =$classical angular frequency.

$E_n=\left(n-\frac{1}{2}\right)\frac{h\left(2\pi f\right)}{2\pi }$

$=\left(n-\frac{1}{2}\right)hf$

$=\left(n-\frac{1}{2}\right)h\left(\frac{c}{\lambda }\right)$

Step 2. Energy of n=1 state

$E_1=\left(1-\frac{1}{2}\right)h\left(\frac{c}{\lambda }\right)$

$=\frac{hc}{2\lambda }$

$E_1=\frac{\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2(5.8\mu \mathrm{m})}$

$=\frac{\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2(5.8\mu \mathrm{m})\left(\frac{{10}^{-6}\mathrm{m}}{1.0\mu \mathrm{m}}\right)}\left(\frac{1.0\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right)$

$=0.107\mathrm{eV}$

Step 2. Energy of n=2 state

$E_2=\left(2-\frac{1}{2}\right)h\left(\frac{c}{\lambda }\right)$

$=\frac{3hc}{2\lambda }$

$E_2=\frac{3\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2(5.8\mu \mathrm{m})}$

$=\frac{3\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2(5.8\mu \mathrm{m})\left(\frac{{10}^{-6}\mathrm{m}}{1.0\mu \mathrm{m}}\right)}\left(\frac{1.0\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right)$

$=0.321\mathrm{eV}$

Step 3. Energy of n=3 state

$E_3=\left(3-\frac{1}{2}\right)h\left(\frac{c}{\lambda }\right)$

$=\frac{5hc}{2\lambda }$

$E_3=\frac{5\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2(5.8\mu \mathrm{m})}$

$=\frac{5\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2(5.8\mu \mathrm{m})\left(\frac{{10}^{-6}\mathrm{m}}{1.0\mu \mathrm{m}}\right)}\left(\frac{1.0\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right)$

$=0.535\mathrm{eV}$