Question

Two adjacent energy levels of an electron in a harmonic potential well are known to be $2.0eV$ and $2.8eV$. What is the spring constant of the potential well?

Short answer

The spring constant for potential well=$1.4N/m$

Step-by-step solution

Step 1. Given information

Energy level of quantum harmonic oscillator =

$E_n=\left(n-\frac{1}{2}\right)\hslash \omega$

Here,

$n=$denotes the state,

$\hslash$= reduced Planck's constant and

$\omega =$classical angular frequency.

Adjacent energy level=

$E_{n+1}=\left((n+1)-\frac{1}{2}\right)\hslash \omega$

$=\left(n+\frac{1}{2}\right)\hslash \omega$

Step 2. For the angular frequency 

$=\hslash \omega$

$\omega =\frac{E_{n+1}-E_n}{\hslash }$

$\omega =\frac{2.8\mathrm{eV}-2.0\mathrm{eV}}{1.05\times {10}^{-34}\mathrm{J}·\mathrm{s}}$

$=\frac{1.28\times {10}^{-19}}{1.05\times {10}^{-34}\mathrm{J}·\mathrm{s}}$

width="126">=1.22×1015s-1

Step 3. For spring constant

$\omega =\sqrt{\frac{k}{m}}$

Here,

$k=$spring constant of the potential well

$m=$electron mass.

$\frac{k}{m}={\omega }^2$

$k=m{\omega }^2$

$k=\left(9.1\times {10}^{-31}\mathrm{kg}\right){\left(1.22\times {10}^{15}{\mathrm{s}}^{-1}\right)}^2$

$=1.354\mathrm{N}/\mathrm{m}\approx 1.4\mathrm{N}/\mathrm{m}$

Thus, the spring constant for potential well =$1.4N/m$