Question

What is the energy of the photon emitted by the harmonic oscillator with stiffness ${\mathbf{k}}_{\mathbf{s}}$ and mass $\mathit{m}$ when it drops from energy level 5 to energy level 2?

Short answer

The energy emitted by the photon is $3\hslash \sqrt{\frac{k_s}{m}}$.

Step-by-step solution

Identification of given data

The mass of harmonic oscillator is$m$.

The stiffness of harmonic oscillator is $k_s$.

The higher energy level for photon is $n_h=5$.

The lower energy level for photon is $n_l=2$.

Conceptual Explanation

The energy of the photon is always in quantized form and energy emitted by photon is equal to the drop in energy between these levels.

Determination of energy emitted by the photon

The angular frequency of the harmonic oscillator is given as:

${\omega }_0=\sqrt{\frac{k_s}{m}}$

The energy of harmonic oscillator for higher level is given as:

$E_h=n_h\hslash {\omega }_0$

The energy of harmonic oscillator for lower level is given as:

$E_l=n_l\hslash {\omega }_0$

The energy emitted by the photonis given as:

$$\begin{gathered} E=E_h-E_l \\ E=n_h\hslash {\omega }_0-n_l\hslash {\omega }_0 \\ E=\left(n_h-n_l\right)\hslash \sqrt{\frac{k_s}{m}} \end{gathered}$$

Here, $\hslash$ is the Planck’s constant.

Substitute all the values in the above equation.

$$\begin{gathered} E=\left(5-2\right)\hslash \sqrt{\frac{k_s}{m}} \\ E=3\hslash \sqrt{\frac{k_s}{m}} \end{gathered}$$

Therefore, the photon emits the energy $3\hslash \sqrt{\frac{k_s}{m}}$ between given levels of harmonic oscillator.