A bottle contains a gas with atoms whose lowest four energy levels are $-\mathbf{12}\mathbf{eV}$, $-\mathbf{6}\mathbf{eV}$, $-\mathbf{3}\mathbf{eV}$, and $-\mathbf{2}\mathbf{eV}$. Electrons run through the bottle and excite the atoms so that at all times there are large numbers of atoms in each of these four energy levels, but there are no atoms in higher energy levels. List the energies of the photons that will be emitted by the gas.

Next, the electron beam is turned off, and all the atoms are in the ground state. Light containing a continuous spectrum of photon energies from $\mathbf{0}.\mathbf{5}\mathbf{eV}$to $\mathbf{15}\mathbf{eV}$shines through the bottle. A photon detector on the other side of the bottle shows that some photon energies are depleted in the spectrum (“dark lines”). What are the energies of the missing photons?

The given data is listed below as,

• The four lowest levels of energy are, $E_0=-12\text{eV}$, $E_1=-6\text{eV}$, $E_2=-3\text{eV}$and $E_3=-2\text{eV}$.

• The continuous spectrum of light is from $0.5\text{eV}$to $15\text{eV}$.

The change in the energy of the photon is equal to the difference in the energy of the higher state and the energy of the lower or ground state.

The equation of the photon energies can be expressed as,

$\Delta E=E_f-E_0$ …(1)

Here, $\Delta E$is the emitted energy of photon, $E_f$is the energy in excited state and $E_0$ is energy in the ground state.

Observing the energy levels, it can be identified that the energy at the ground level, first, second and the third excited level is $-12\text{eV}$, $-6\text{eV}$, $-3\text{eV}$and $-2\text{eV}$respectively.

For the electrons going from the ground state to the first excited state,

For $E_f=E_1=-6\text{eV}$and $E_0=-12\text{eV}$in equation (1).

$$\begin{gathered} △E=-6eV-(-12eV) \\ =6eV \end{gathered}$$

For the electrons going from the ground state to the second excited state,

For $E_f=E_2=-3\text{eV}$and$E_0=-12\text{eV}$in equation (1).

$$\begin{gathered} △E=-3eV-(-12eV) \\ =9eV \end{gathered}$$

For the electrons going from the ground state to the third excited state,

For $E_f=E_3=-2\text{eV}$and$E_0=-12\text{eV}$in equation (1).

$$\begin{gathered} △E=-2eV-(-12eV) \\ =10eV \end{gathered}$$

For the electrons going from the first excited state to the second excited state, the equation becomes,

$\Delta E=E_f-E_1$ …(2)

Here, $\Delta E$is the energy emitted by the photon, $E_f$is the energy of the other excited state and $E_1$is the energy of the first excited state.

For the electrons going from the first excited state to the second excited state

For $E_f=E_2=-3\text{eV}$and$E_1=-6\text{eV}$in equation (2).

$$\begin{gathered} △E=-3eV-(-6eV) \\ =3eV \end{gathered}$$

For the electrons going from the first excited state to the third excited state.

For $E_f=E_3=-2\text{eV}$and $E_1=-6\text{eV}$in equation (2).

$$\begin{gathered} △E=-2eV-(-6eV) \\ =4eV \end{gathered}$$

For the electrons going from the second excited state to the third excited state, the equation becomes,

$\Delta E=E_f-E_2$ …(3)

Here, $\Delta E$is the energy emitted by the photon, $E_f$is the energy of the other excited state and $E_2$is the energy of the second excited state.

For $E_f=E_3=-2\text{eV}$and$E_2=-3\text{eV}$in equation (3).

$$\begin{gathered} △E=-2eV-(-3eV) \\ =1eV \end{gathered}$$

During the transition of the electrons, the electrons are not available in the excited state. During the energy strike, the photon detector also shows that the dark lines which are the depleted photon energies. Hence, the missing dark lines are the transition of the energy between the ground state to the first, second and the third excited state which are $6\text{eV}$, $9\text{eV}$, $10\text{eV}$ respectively.

Thus, the energies of photon that will be emitted by the gas are $6\text{eV}$, $9\text{eV}$, $10\text{eV}$, $3\text{eV}$, $4\text{eV}$and $1\text{eV}$and the energies of the missing photons are $6\text{eV}$, $9\text{eV}$, $10\text{eV}$.