 Agaiepsh

2021-11-15

The energy of a vibrating molecule is quantized much like the energy of an electron in the hydrogen atom. The energy levels of a vibrating molecule are given by the equation: ${E}_{n}=\left(n+\frac{1}{2}hv\right)$ where n is a quantum number with possible values of 1, 2, ... and v is the frequency of vibration. The vibration frequency of HCI is $8.85×{10}^{13}{s}^{-1}$ What minimum energy is required to excite a vibration in HCl? What wavelength of light is required to excite this vibration? Linda Tincher

Expert

Step 1
A vibrating molecule's energy is ${E}_{n}=\left(n+\frac{1}{2}\right)hv$ when the quantum number is n; $h=6.626×{10}^{-34}j\cdot s$ and v is the vibration frequency.
Given that the vibration frequency of HCl is $8.85×{10}^{13}s$
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 $\mathrm{△}E=\frac{hc}{\lambda }$ to find the vibration's corresponding wavelength where $c=3.00×{10}^{8}\frac{m}{s}$
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 $n=1\to n=2$. Calculate
${E}_{1}=\left(n+\frac{1}{2}\right)hv=\left(1+\frac{1}{2}\right)\cdot 6.626×{10}^{-34}j\cdot s\cdot 8.85×{10}^{13}s=8.796015×{10}^{-20}j$
${E}_{2}=\left(n+\frac{1}{2}\right)hv=\left(2+\frac{1}{2}\right)\cdot 6.626×{10}^{-34}j\cdot s\cdot 8.85×{10}^{13}s=1.466×{10}^{-19}j$
Step 3
Calculate
$\mathrm{△}{E}_{1}\to 2={E}_{f}-{E}_{i}$
$\mathrm{△}{E}_{1}\to 2={E}_{2}-{E}_{1}=1.466×{10}^{-19}j-8.796×{10}^{-20}j=\begin{array}{|c|}\hline 5.864×{10}^{-20}J\\ \hline\end{array}$
Step 4
Calculate $\lambda$ using $\mathrm{△}E=\frac{hc}{\lambda }$
${\lambda }_{1}\to 2\frac{\left(6.626×{10}^{-}{34}_{J}\cdot s\right)\cdot \left(3.00×{10}^{8}m/s\right)}{5.864×{10}^{-20}J}=3.3898×{10}^{-6}m\cdot \frac{1×{10}^{9}nm}{1m}=\begin{array}{|c|}\hline 3390nm\\ \hline\end{array}$
$\mathrm{△}{E}_{1}\to 2=5.864×{10}^{-20}j$
${\lambda }_{1}\to 2=3390nm$

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