Quantum Chemistry/Example 16

Write an example question showing the calculation of the frequency of EM radiation emitted when a HCl molecule transitions from the J=1–>0 rotational state.

Example 16: Giving the bond length r_e=1.27Å, find the frequency of the EM radiation emitted when a HCl molecule transitions from the J=1→0 rotational state

Given the reduced mass (μ) and inertia (I), the energy can be found and subsequently be used to find frequency of the transition.

Note that for rotational states for J=1→0, ΔE is not required as the transition states are in the ground state.

Inertia can be found by using reduced mass of the molecule and their bond length(r_e = 1.27Å):

$I=\mu \cdot r_e^2$

where reduced mass can be calculated by the following:

$\mu =\frac{m_1\cdot m_2}{m_1+m_2}$, where $m_1=1.00784\text{amu}$ and $m_2=35.453\text{amu}$

$\mu =\frac{(1.00784)(35.453)}{1.00784+35.453}=0.9770\text{amu}$

Convert to SI units:

$1u=1.66054\times 10^{-27}kg$, thus

$\mu =0.9799\text{amu}\times \frac{1.66054\times 10^{-27}kg}{1\text{amu}}=1.627\times 10^{-27}kg$

Note that the bond length must also be in SI units,

1Å= 1×10^-10m

Bond length = 1.27×10^-10m

Inertia can now be calculated:

$I=(1.627\times 10^{-27}kg)(1.27\times 10^{-10}m{)}^2=2.624\times 10^{-47}kg\cdot m^2$

After calculating the inertia, energy can be found:

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

and $\hslash =\frac{h}{2\pi }$, where h is the Planck's constant.

$E=(\frac{6.626\times 10^{-34}{m}^{2}kg{s}^{-1}}{2\pi }{)}^2\times \frac{1}{2\times 2.624\times 10^{-47}kg\cdot m^2}\times 1(1+1)$

$E=4.240\times 10^{-22}J$

Frequency of the transmission can therefore be found:

$\nu =\frac{E}{h}$

$\nu =\frac{4.240\times 10^{-22}J}{6.626\times 10^{-34}{m}^{2}kg{s}^{-1}}=6.40\times 10^{11}{s}^{-1}$