The energy of a vibrating molecule is quantized much like the energy of an electron...

Step 1
A vibrating molecule's energy is ${\displaystyle E_n=(n+\frac{1}{2})hv}$ when the quantum number is n; ${\displaystyle h=6.626\times {10}^{-34}j\cdot s}$ and v is the vibration frequency.
Given that the vibration frequency of HCl is ${\displaystyle 8.85\times {10}^{13}s}$
Find the wavelength of light that can trigger this vibration and the minimal energy needed to cause HCl to vibrate.
Solution: The difference between the energies of the two lowest quantum numbers should represent the minimal energy needed to cause HCl to vibrate. Use ${\displaystyle \mathrm{△}E=\frac{hc}{\lambda }}$ to find the vibration's corresponding wavelength where ${\displaystyle c=3.00\times {10}^8\frac{m}{s}}$
Step 2
Since n should be a positive number in a quantum system, the transition that corresponds to the minimal amount of energy needed to for HCl to vibrate is ${\displaystyle n=1\to n=2}$. Calculate ${\displaystyle E_1\text{ and }{E}_2}$
${\displaystyle E_1=(n+\frac{1}{2})hv=(1+\frac{1}{2})\cdot 6.626\times {10}^{-34}j\cdot s\cdot 8.85\times {10}^{13}s=8.796015\times {10}^{-20}j}$
${\displaystyle E_2=(n+\frac{1}{2})hv=(2+\frac{1}{2})\cdot 6.626\times {10}^{-34}j\cdot s\cdot 8.85\times {10}^{13}s=1.466\times {10}^{-19}j}$
Step 3
Calculate
${\displaystyle \mathrm{△}{E}_1\to 2=E_f-E_i}$
$\mathrm{△}{E}_1\to 2=E_2-E_1=1.466\times {10}^{-19}j-8.796\times {10}^{-20}j=\begin{array}{|c|}\hline 5.864\times {10}^{-20}J\\ \hline\end{array}$
Step 4
Calculate ${\displaystyle \lambda }$ using ${\displaystyle \mathrm{△}E=\frac{hc}{\lambda }}$
${\lambda }_1\to 2\frac{(6.626\times {10}^{-}{34}_J\cdot s)\cdot (3.00\times {10}^{8}m/s)}{5.864\times {10}^{-20}J}=3.3898\times {10}^{-6}m\cdot \frac{1\times {10}^{9}nm}{1m}=\begin{array}{|c|}\hline 3390nm\\ \hline\end{array}$
${\displaystyle \mathrm{△}{E}_1\to 2=5.864\times {10}^{-20}j}$
${\displaystyle {\lambda }_1\to 2=3390nm}$