# Quantum Chemistry/Example 17

The Bond 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 of HCl(K), 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

$\mu ={\frac {1.007842u\cdot 35.453u}{1.007842u+35.453u}}=0.979982u$

convert to the SI unit of Kg

$1u=1.66054\cdot 10^{-27}Kg$

$0.979983u=0.979983u\cdot 1.66054\cdot 10^{-27}Kg$

$\mu =1.6273\cdot 10^{-27}Kg$

To find the fundermental frequency

$\nu _{0}={\frac {1}{2\pi }}\left({\frac {K}{\mu }}\right)^{1/2}$

$={\frac {1}{2\pi }}\left({\frac {480N/m}{1.6273\cdot 10^{-27}Kg}}\right)^{1/2}$
$=8.6438\cdot 10^{13}hz$

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

$\Delta E=hv$
$hv=h\nu _{0}$
$v=\nu _{0}=8.6438\cdot 10^{13}hz$