Question

An electron in a harmonic potential well absorbs a photon with a wavelength of $400nm$as it undergoes a $1\to 2$ quantum jump. What wavelength is absorbed in a $1\to 3$ quantum jump?

Short answer

Wavelength absorbed in $1\to 2$quantum jump = $200nm$

Wavelength absorbed in $1\to 3$quantum jump=$200nm$

Step-by-step solution

Step 1. Given information

Energy level of harmonic quantum oscillator,

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

Here,

n denotes the state,

$h=$Planck's constant and

$\omega =$ classical angular frequency.

Step 2. Quantum jump between 1→2

$E_1=\left(1-\frac{1}{2}\right)\frac{h\omega }{2\pi }$

$=\frac{1}{2}\left(\frac{h\omega }{2\pi }\right)$

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

$=\frac{3}{2}\left(\frac{h\omega }{2\pi }\right)$

Now,

${\lambda }_{\text{photon,}1\to 2}=\frac{hc}{\frac{3}{2}\left(\frac{h\omega }{2\pi }\right)-\frac{1}{2}\left(\frac{h\omega }{2\pi }\right)}$

$=\frac{hc}{\left(\frac{h\omega }{2\pi }\right)}$

$=\frac{2\pi c}{\omega }$

$\frac{2\pi c}{\omega }=400\mathrm{nm}$

$\frac{\pi c}{\omega }=200\mathrm{nm}$

Step 3. Quantum jump between 1→3

$E_1=\frac{1}{2}\left(\frac{h\omega }{2\pi }\right)$

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

$=\frac{5}{2}\left(\frac{h\omega }{2\pi }\right)$

${\lambda }_{\text{photon,}1\to 3}=\frac{hc}{\frac{5}{2}\left(\frac{h\omega }{2\pi }\right)-\frac{1}{2}\left(\frac{h\omega }{2\pi }\right)}$

$=\frac{hc}{\left(\frac{4}{2}\right)\left(\frac{h\omega }{2\pi }\right)}$

$=\frac{\pi c}{\omega }$

As found earlier.

$\frac{\pi c}{\omega }=200\mathrm{nm}$

${\lambda }_{\text{photon,}1\to 3}=200\mathrm{nm}$

Thus, the wavelength absorbed in$1\to 3$quantum jump =$200nm$