Consider a microscopic spring–mass system whose spring stiffness is$50\mathrm{N}/\mathrm{m}$, and the mass is$4\times {10}^{-26}\mathrm{kg}$. (a) What is the smallest amount of vibrational energy that can be added to this system? (b) What is the difference in mass (if any) of the microscopic oscillator between being in the ground state and being in the first excited state? (c) In a collection of these microscopic oscillators, the temperature is high enough that the ground state and the first three excited states are occupied. What are possible energies of photons emitted by these oscillators?

The given data can be listed below as-

• The spring stiffness of the microscopic spring-mass system is $50\mathrm{N}/\mathrm{m}.$
• The mass of the spring is$4\times {10}^{-26}\mathrm{kg}.$

The Planck’s law states that the energy emitted is directly proportional to the product of the Planck’s constant and the frequency.

The mass-energy equivalence states that the energy emitted is equal to the mass of an object and the square of the light’s speed.

The Planck’s equation gives the vibrational energy and the mass energy equivalence gives the difference in the mass and the possible photon energies.

According to the Planck’s law, the smallest vibrational energy is expressed as-

$∆E=h{\omega }_0$

Here, $∆E=$The smallest amount of vibrational energy

$h=$ The Planck’s constant’s

${\omega }_0=$The angular frequency of the spring mass system

The above equation can also be written as-

$∆E=\left(\frac{h}{2\mathrm{\pi }}\right).\left(\sqrt{\frac{k}{m}}\right)$

Here, $h=$Absolute value of the Planck’s constant

$k=$The spring stiffness

$m=$Mass of the spring

Forrole="math" localid="1657799146287" $\mathrm{h}=6.626\times {10}^{-34}\mathrm{Js},k=50\mathrm{N}/\mathrm{m}\mathrm{and}4\times {10}^{-26}\mathrm{kg},$

$$\begin{gathered} ∆E=\left(\frac{6.626\times {10}^{-34}\mathrm{Js}}{2\times 3.14}\right).\sqrt{\left(\frac{50\mathrm{N}/\mathrm{m}}{4\times {10}^{-26}kg}\right)} \\ =1.05\times {10}^{-34}\mathrm{Js}.3.53\times {10}^{13}{\mathrm{s}}^{-1}(\mathrm{N}/\mathrm{kgm}=\mathrm{N}/\mathrm{Ns})(\therefore 1\mathrm{Ns}=1\mathrm{kgm}) \\ =3.71\times {10}^{-21}\mathrm{J} \end{gathered}$$

Thus, the smallest amount of vibrational energy that can be added to this system is $3.71\times {10}^{-21}\mathrm{J}$.

According to the mass-energy equivalence, the equation of the energy of the microscopic oscillator is expressed as-

$∆E=∆m{c}^2$

Here,$∆E=$ Energy of the microscopic oscillator

$∆m=$Mass difference of the microscopic oscillator

$c=$The speed of light

For and$∆E=3.71\times {10}^{-21}\mathrm{J}\mathrm{and}\mathrm{c}=3\times {10}^8\mathrm{m}/{\mathrm{s}}^2.$ ,

$$\begin{gathered} 3.71\times {10}^{-21}\mathrm{J}=\mathrm{m}\times {(3\times {10}^8\mathrm{m}/\mathrm{s})}^2\mathrm{m}=\frac{3.71\times {10}^{-21}\mathrm{J}}{{(3\times {10}^8\mathrm{m}/\mathrm{s})}^2} \\ =4.1\times {10}^{-38}{\mathrm{Js}}^2/{\mathrm{m}}^2\times {\mathrm{kgm}}^2/{\mathrm{s}}^2 \\ =4.1\times {10}^{-38}\mathrm{kg} \end{gathered}$$

Thus, the difference in mass between being in the ground and the first exited state is $4.1\times {10}^{-38}\mathrm{kg}$.

As the energy levels are at equal distance, then there are mainly three possible energies of photon.

According to the Planck’s law, the equation of the energy of the photon at the ground state is expressed as-

$∆E_1=NE$

Here,$E_1$the energy of the photon at the ground state

$N=$The number of the state

$E=$The energy level

For $N=1$and$∆\mathrm{E}=3.71\times {10}^{-21}\mathrm{J}$

$$\begin{gathered} ∆\mathrm{E}=1\times 3.71\times {10}^{-21}\mathrm{J} \\ =3.71\times {10}^{-21}\mathrm{J} \end{gathered}$$

According to the Planck’s law, the equation of the energy of the photon at the first exited state is expressed as-

$∆E_2=NE$

Here, $∆E_2=$the energy of the photon at the first exited state

$N=$The number of the state

$E=$The energy level

For$\mathrm{N}=2\mathrm{and}∆\mathrm{E}=3.71\times {10}^{-21}\mathrm{J}$.

$$\begin{gathered} ∆{\mathrm{E}}_1=2\times 3.71\times {10}^{-21}\mathrm{J} \\ =7.42\times {10}^{-21}\mathrm{J} \end{gathered}$$

According to the Planck’s law, the equation of the energy of the photon at the second excited state is expressed as-

$∆E_2=NE$

Here,$∆E_2=$the energy of the photon at the second excited state

$N=$The number of the state

$E=$The energy level

For $\mathrm{N}=3\mathrm{and}∆\mathrm{E}=3.71\times {10}^{-21}\mathrm{J}$

$$\begin{gathered} ∆\mathrm{E}=3\times 3.71\times {10}^{-21}\mathrm{J} \\ =11.13\times {10}^{-21}\mathrm{J} \end{gathered}$$

Thus, the possible energies of photons emitted by these oscillators are$3.71\times {10}^{-21}\mathrm{J}$ , $7.42\times {10}^{-21}\mathrm{J}$and$11.13\times {10}^{-21}\mathrm{J}$ respectively.