Quantum Chemistry/Example 17

The force constant of HCl is determined computationally to be 480 N/m. Given this information find the frequency of EM radiation required to excite the HCl molecule from its ground state to its first excited state.

Solution
Given the bond constant (k) of HCl, we use the relationship between fundamental frequency and bond constant to find the bond constant.

$$ \nu_0={\frac{1}{2\pi}} {\left( {\frac{k}{\mu}} \right)}^{1/2} $$

where $$K$$ is the bond constant.


 * $$\mu$$ is the reduced mass of HCl.

To find the reduced mass of HCl the masses of H and Cl are multiplied and divided by the sum of the masses. $$ \mu={\frac{m_1\cdot m_2}{m_1+m_2}} $$ For HCl the reduced mass is calculated as $$ \mu={\frac{1.007842u\cdot35.453u}{1.007842u+35.453u}}

=0.979982u $$ convert to the SI unit of Kg  $$1u=1.66054\cdot10^{-27}Kg $$ $$0.979983u= 0.979983u\cdot1.66054 \cdot 10^{-27}Kg$$ $$\mu= 1.6273\cdot10^{-27}Kg$$ To find the fundamental frequency $$ \nu_0={\frac{1}{2\pi}}\left(\frac{K}{\mu} \right)^{1/2} $$
 * $$= {\frac{1}{2\pi}}\left(\frac{480N/m}{1.6273\cdot10^{-27}Kg} \right)^{1/2}$$
 * $$= 8.6438\cdot 10^{13} hz $$

After finding the fundamental frequency, the Energy at different quantum levels can be found by $$E_v=h\nu_0 \left(v + \frac{1}{2} \right)$$ For the ground state i.e. $$v=0$$ $$E_0=h\nu_0 \left(0 + \frac{1}{2} \right)$$ $$E_0=h\nu_0 \left(\frac{1}{2}\right)$$ For the first excited state i.e. $$v=1$$ $$E_1=h\nu_0 \left(1 + \frac{1}{2} \right)$$ $$E_1=h\nu_0 \left(\frac{3}{2}\right)$$ The difference in energy between the two states is $$\Delta E=E_1-E_0$$ $$\Delta E=h\nu_0 \left(\frac{3}{2}\right) - h\nu_0 \left(\frac{1}{2}\right)$$ $$\Delta E=h\nu_0$$ and Energy is defined as Planck's constant multiplied by frequency $$\Delta E=hv$$ $$hv=h\nu_0$$ $$v=\nu_0=8.6438\cdot 10^{13} hz$$