Agaiepsh

Answered

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: ${E}_{n}=(n+\frac{1}{2}hv)$ 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 $8.85\times {10}^{13}{s}^{-1}$ What minimum energy is required to excite a vibration in HCl? What wavelength of light is required to excite this vibration?

Answer & Explanation

Linda Tincher

Expert

2021-11-16Added 14 answers

Step 1

A vibrating molecule's energy is ${E}_{n}=(n+\frac{1}{2})hv$ when the quantum number is n; $h=6.626\times {10}^{-34}j\cdot s$ and v is the vibration frequency.

Given that the vibration frequency of HCl is $8.85\times {10}^{13}s$

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 $\mathrm{\u25b3}E=\frac{hc}{\lambda}$ to find the vibration's corresponding wavelength where $c=3.00\times {10}^{8}\frac{m}{s}$

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 $n=1\to n=2$. Calculate $E}_{1}\text{}\text{and}\text{}{E}_{2$

${E}_{1}=(n+\frac{1}{2})hv=(1+\frac{1}{2})\cdot 6.626\times {10}^{-34}j\cdot s\cdot 8.85\times {10}^{13}s=8.796015\times {10}^{-20}j$

${E}_{2}=(n+\frac{1}{2})hv=(2+\frac{1}{2})\cdot 6.626\times {10}^{-34}j\cdot s\cdot 8.85\times {10}^{13}s=1.466\times {10}^{-19}j$

Step 3

Calculate

$\mathrm{\u25b3}{E}_{1}\to 2={E}_{f}-{E}_{i}$

$\mathrm{\u25b3}{E}_{1}\to 2={E}_{2}-{E}_{1}=1.466\times {10}^{-19}j-8.796\times {10}^{-20}j=\begin{array}{|c|}\hline 5.864\times {10}^{-20}J\\ \hline\end{array}$

Step 4

Calculate $\lambda$ using $\mathrm{\u25b3}E=\frac{hc}{\lambda}$

${\lambda}_{1}\to 2\frac{(6.626\times {10}^{-}{34}_{J}\cdot s)\cdot (3.00\times {10}^{8}m/s)}{5.864\times {10}^{-20}J}=3.3898\times {10}^{-6}m\cdot \frac{1\times {10}^{9}nm}{1m}=\begin{array}{|c|}\hline 3390nm\\ \hline\end{array}$

$\mathrm{\u25b3}{E}_{1}\to 2=5.864\times {10}^{-20}j$

${\lambda}_{1}\to 2=3390nm$

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