Question

Suppose that a collection of quantum harmonic oscillators occupies the lowest four energy levels, and the spacing between levels is $\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{eV}$. What is the complete emission spectrum for this system? That is, what photon energies will appear in the emissions? Include all energies, whether or not they fall in the visible region of the electromagnetic spectrum.

Short answer

The photon energies for the given system are$0.4\mathrm{eV}$ (three transitions), $0.4\mathrm{eV}$(Two transitions), and$1.2\mathrm{eV}$ (one transition).

Step-by-step solution

Concept Introduction

The expression for the energy of the oscillator is given by,

${\mathbf{E}}_{\mathbf{N}}\mathbf{=}{\mathbf{Nh\omega }}_{\mathbf{0}}\mathbf{+}{\mathbf{E}}_{\mathbf{0}}$

Here $\mathrm{N}=0,1,2,\mathrm{....}$

The angular frequency is given by,

${\mathrm{\omega }}_0=\sqrt{\frac{{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}$

Here${\mathrm{\omega }}_0$ is the angular frequency,${\mathrm{k}}_{\mathrm{s}}$ is the spring constant,$\mathrm{m}$ is the mass.

The ground state energy of the harmonic oscillator is given by,

${\mathrm{E}}_0=\frac{1}{2}{\mathrm{h\omega }}_0$

Given data

The spacing between the energy level is given by,

$\mathrm{\Delta E}=0.4\mathrm{eV}$

${\mathrm{h\omega }}_0=0.4\mathrm{eV}$

Calculate the photon energy that will appear in the emissions.

The expression for energy of the photon which is emitted during the transition from $3\to 2$is given by,

$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_2=(3{\mathrm{h\omega }}_0+{\mathrm{E}}_0)-(2{\mathrm{h\omega }}_0+{\mathrm{E}}_0)\\ ={\mathrm{h\omega }}_0\end{array}$

Substitute$0.4\mathrm{eV}$ for ${\mathrm{h\omega }}_0$into the above equation,

${\mathrm{E}}_3-{\mathrm{E}}_2=0.4\mathrm{eV}$

The expression for energy of the photon which is emitted during the transition from$3\to 0$is given by,

$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_0=(3{\mathrm{h\omega }}_0+{\mathrm{E}}_0)-\left({\mathrm{E}}_0\right)\\ =3{\mathrm{h\omega }}_0\end{array}$

Substitute$0.4\mathrm{eV}$ for ${\mathrm{h\omega }}_0$into the above equation,

$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_0=3\left(0.4\mathrm{eV}\right)\\ =1.2\mathrm{eV}\end{array}$

The expression for energy of the photon which is emitted during the transition from$2\to 0$is given by,

$\begin{array}{c}{\mathrm{E}}_2-{\mathrm{E}}_0=(2{\mathrm{h\omega }}_0+{\mathrm{E}}_0)-\left({\mathrm{E}}_0\right)\\ =2{\mathrm{h\omega }}_0\end{array}$

Substitute$0.4\mathrm{eV}$for ${\mathrm{h\omega }}_0$into the above equation,

$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_0=2\left(0.4\mathrm{eV}\right)\\ =0.8\mathrm{eV}\end{array}$

The expression for energy of the photon which is emitted during the transition from$3\to 1$is given by,

$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_1=(3{\mathrm{h\omega }}_0+{\mathrm{E}}_0)-({\mathrm{h\omega }}_0+{\mathrm{E}}_0)\\ =2{\mathrm{h\omega }}_0\end{array}$

Substitute $0.4\mathrm{eV}$ for ${\mathrm{h\omega }}_0$into the above equation,

$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_0=2\left(0.4\mathrm{eV}\right)\\ =0.8\mathrm{eV}\end{array}$

The expression for energy of the photon which is emitted during the transition from$1\to 0$is given by,

$\begin{array}{c}{\mathrm{E}}_1-{\mathrm{E}}_0=({\mathrm{h\omega }}_0+{\mathrm{E}}_0)-\left({\mathrm{E}}_0\right)\\ ={\mathrm{h\omega }}_0\end{array}$

Substitute$0.4\mathrm{eV}$ for ${\mathrm{h\omega }}_0$into the above equation,

${\mathrm{E}}_3-{\mathrm{E}}_0=0.4\mathrm{eV}$

Therefore, the photon energies for the given system are $0.4\mathrm{eV}$(three transitions),$0.4\mathrm{eV}$ (Two transitions), and $1.2\mathrm{eV}$(one transition).