Question

Assume that a hypothetical object has just four quantum states, with the following energies:

$-\mathbf{1}.\mathbf{0}\mathbf{eV}$(third excited state)

$-\mathbf{1}.\mathbf{8}\mathbf{eV}$(second excited state)

$-\mathbf{2}.\mathbf{9}\mathbf{eV}$(first excited state)

$-\mathbf{4}.\mathbf{8}\mathbf{eV}$(ground state)

(a) Suppose that material containing many such objects is hit with a beam of energetic electrons, which ensures that there are always some objects in all of these states. What are the six 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.) List the photon emission energies. (b) Next, suppose that the beam of electrons is shut off so that all of the objects are in the ground state almost all the time. If electromagnetic radiation with a wide range of energies is passed through the material, what will be the three energies of photons corresponding to missing (“dark”) lines in the spectrum? Remember that there is hardly any absorption from excited states, because emission from an excited state happens very quickly, so there is never a significant number of objects in an excited state. Assume that the detector is sensitive to a wide range of photon energies, not just energies in the visible region. List the dark-line energies.

Short answer

(a) $0.8\text{eV}$, $1.9\text{eV}$,$1.1\text{eV}$, $3.8\text{eV}$, $3.0\text{eV}$, and $1.9\text{eV}$

(b) $1.9\text{eV}$, $3.0\text{eV}$, $3.8\text{eV}$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The energy in the third excited state is, $E_3=-1.0\text{eV}$
• The energy in the second excited state is, $E_2=-1.8\text{eV}$
• The energy in the first excited state is, $E_1=-2.9\text{eV}$
• The energy in the ground state is,

$E_0=-4.8\text{eV}$

Significance of the change in the photon energies

The change in the photon energies is equal to the difference between the energy in the higher state and the energy in the 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.

Determination of the emission energy of photon

(a)

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

For $E_f=E_1=-2.9\text{eV}$and $E_0=-4.8\text{eV}$in equation (1).

$$\begin{gathered} △E=-2.9eV-(-4.8eV) \\ =1.9eV \end{gathered}$$

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

For $E_f=E_2=-1.8\text{eV}$and $E_0=-4.8\text{eV}$in equation (1).

$$\begin{gathered} △E=-1.8eV-(-4.8eV) \\ =3eV \end{gathered}$$

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

For $E_f=E_3=-1.0\text{eV}$and $E_0=-4.8\text{eV}$in equation (1).

$$\begin{gathered} △E=-1.0eV-(-4.8eV) \\ =3.8eV \end{gathered}$$

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 $E_f=E_2=-1.8\text{eV}$and $E_1=-2.9\text{eV}$in equation (2).

$$\begin{gathered} △E=-1.8eV-(-2.9eV) \\ =1.1eV \end{gathered}$$

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

For $E_f=E_3=-1.0\text{eV}$and$E_1=-2.9\text{eV}$in equation (2).

$$\begin{gathered} △E=-1.0eV-(-2.9eV) \\ =1.9eV \end{gathered}$$

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 $E_f=E_3=-1.0\text{eV}$and$E_2=-1.8\text{eV}$in equation (3).

$$\begin{gathered} △E=-1.0eV-(-1.8eV) \\ =0.8eV \end{gathered}$$

Thus, the six energies of photon that could be strongly emitted by the material or the list of the photon emission energies are $1.9\text{eV}$, $3\text{eV}$, $3.8\text{eV}$, $1.1\text{eV}$, $1.9\text{eV}$and $0.8\text{eV}$.

Determination of the energy of dark lines

(b)

The dark lines in the spectrum indicates that no electrons are available in the exited state due to the possible transitions of the electrons. However, when a large amount of energy strikes, the energy frequencies are absorbed as the black bands inside the spectrum. 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 $1.9\text{eV}$, $3\text{eV}$, $3.8\text{eV}$respectively.

Thus, the missing lines or the dark-line energies in the spectrum are $1.9\text{eV}$, $3\text{eV}$, $3.8\text{eV}$.