Question

An electron confined in a harmonic potential well emits a 1200nm photon as it undergoes a $3\to 2$quantum jump. What is the spring constant of the potential well?

Short answer

Spring constant of potential well =$2.243N/m$

Step-by-step solution

Step 1. Given information

Energy levels 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.

Step 2. Energy of second level and third level

for second level,

$E_2=\left(2-\frac{1}{2}\right)\hslash \omega$

$=\frac{3}{2}\hslash \omega$

for third level,

$E_3=\left(3-\frac{1}{2}\right)\hslash \omega$

$=\frac{5}{2}\hslash \omega$

$\Delta E_{3\to 2}=E_3-E_2$

$=\frac{5}{2}\hslash \omega -\frac{3}{2}\hslash \omega$

$=\hslash \omega$

Step 3. For classical angular frequency

$E_{\text{photon}}=\frac{hc}{{\lambda }_{\text{photon}}}$

$\Delta E_{3\to 2}=\frac{hc}{{\lambda }_{\text{photon}}}$

therefore,

$h\omega =\frac{hc}{{\lambda }_{\text{photon}}}$

$\frac{h\omega }{2\pi }=\frac{hc}{{\lambda }_{\text{photon}}}$

$\omega =\frac{2\pi c}{{\lambda }_{\text{photon}}}$

$\omega =\frac{\left(2\pi \right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{1200\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1.0\mathrm{nm}}\right)}$

$=\frac{18.84\times {10}^8\mathrm{m}/\mathrm{s}}{1.2\times {10}^{-6}\mathrm{m}}$

$=15.7\times {10}^{14}{\mathrm{s}}^{-1}$

Step 4. Finding spring constant

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

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

$k=m{\omega }^2$

$k=\left(9.1\times {10}^{-31}\mathrm{kg}\right){\left(2.5\times {10}^{14}{\mathrm{s}}^{-1}\right)}^2$

$=2.243\mathrm{N}/\mathrm{m}$

Thus, the spring constant of potential well$=2.243\mathrm{N}/\mathrm{m}$