91影视

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,

$螖E=E_f-E_0$ 鈥�(1)

Here, $螖E$ 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).

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

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).

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

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).

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

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

$螖E=E_f-E_1$ 鈥�(2)

Here, $螖E$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),

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

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),

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

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

$螖E=E_f-E_2$ 鈥�(3)

Here, $螖E$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).

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

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.