Assume that a hypothetical object has just four quantum states, with the energies shown in Figure 8.43.

(a) Suppose that the temperature is high enough that in a material containing many such objects, at any instant some objects are found in all of these states. What are all the energies of photons that could be strongly emitted by the material? (In actual quantum objects there are often “selection rules” that forbid certain emissions even though there is enough energy; assume that there are no such restrictions here.) (b) If the temperature is very low and electromagnetic radiation with a wide range of energies is passed through the material, what will be the energies of photons corresponding to missing (“dark”) lines in the spectrum? (Assume that the detector is sensitive to a wide range of photon energies, not just energies in the visible region.)

The given data is listed below as-

• The energy at the ground state is, $E_0$
• The energy at the first excited state is, $E_1=1.9\text{eV}+E_0$
• The energy at the second excited state is,$E_2=2.5\text{eV}+E_0$
• The energy at the third excited state is, $E_3=3.0\text{eV}+E_0$

The change in the energy of the photon equals the difference between the energy at the higher and the energy at the lower 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.

$E_1=1.9\text{eV}$(a)

As there is no energy at the ground state, then the energy at the ground state is zero that is $E_0=0$.

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

For $E_f=E_1=1.9\text{eV}$and $E_0=0\text{eV}$in equation (1).

$$\begin{gathered} △E=1.9eV-\left(0eV\right) \\ =1.9eV \end{gathered}$$

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

For $E_f=E_2=2.5\text{eV}$and $E_0=0\text{eV}$in equation (1).

$$\begin{gathered} △E=2.5eV-\left(0eV\right) \\ =2.5eV \end{gathered}$$

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

For $E_f=E_3=3.0\text{eV}$and $E_0=0\text{eV}$in equation (1).

$$\begin{gathered} △E=3.0eV-\left(0eV\right) \\ =3.0eV \end{gathered}$$

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 $E_f=E_2=2.5\text{eV}$and $E_1=1.9\text{eV}$in equation (2),

$$\begin{gathered} △E=2.5eV-(1.9eV) \\ =0.6eV \end{gathered}$$

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

For $E_f=E_3=3\text{eV}$and $E_1=1.9\text{eV}$in equation (2),

$$\begin{gathered} △E=3eV-(1.9eV) \\ =1.1eV \end{gathered}$$

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 $E_f=E_3=3.0\text{eV}$and $E_2=2.5\text{eV}$in equation (3).

$$\begin{gathered} △E=3.0eV-(2.5eV) \\ =0.5eV \end{gathered}$$

Thus, the energies of photon that could be strongly emitted by the material are $1.9\text{eV}$, $2.5\text{eV}$, $3\text{eV}$, $0.6\text{eV}$, $1.1\text{eV}$and $0.5\text{eV}$.

(b)

The electrons move from the ground to the higher excited state and gain energy. Hence, due to low temperature and the electromagnetic radiation, wide energy range has been passed through the material. So, energy of the photon that is corresponding to the dark lines is from the ground to the higher excited states.

Thus, the energies of the photons that corresponds to the missing lines in the spectrum are$1.9\text{eV}$, $2.5\text{eV}$and $3\text{eV}$respectively.