Question

A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of 0.85 eV makes a transition to a state with an excitation energy (the difference between the energy of the state and that of the ground state) of $\mathbf{10}\mathbf{.}\mathbf{2}\mathit{e}\mathit{V}$. (a) What is the energy of the photon emitted as a result of the transition? What are the (b) higher quantum number and (c) lower quantum number of the transition producing this emission?

Short answer

a) The energy of the photon emitted is .

b) The higher quantum number is .

c) The lower quantum number is .

Step-by-step solution

Describe the expression for the energy of the hydrogen atom for the nth state.

The expression for energy of the hydrogen atom for the ${\mathbf{n}}^{\mathbf{th}}$state is given by,

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\left(-13.6\mathrm{eV}\right)\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}$

The expression for the energy of the hydrogen atom for the lower excited state is given by,

${\mathbf{E}}_{\mathbf{1}}\mathbf{=}\left(-13.6\mathrm{eV}\right)\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}}$ ….. (1)

The expression for the energy of the hydrogen atom for the higher excited state is given by,

${\mathbf{E}}_{\mathbf{2}}\mathbf{=}\left(-13.6\mathrm{eV}\right)\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}$ ….. (2)

(a) Define the energy of the photon emitted as a result of the transition:

Given that, the excitation energy is 10.2 eV and the ground state energy of the hydrogen atom is -13.6 eV. The initial energy of the hydrogen atom is equal to the sum of the energy of the ground state energy and the excitation energy.

$$\begin{gathered} {\mathrm{E}}_1=-13.6\mathrm{eV}+10.2\mathrm{eV} \\ =-3.4\mathrm{eV} \end{gathered}$$

The difference between the final energy of the atom and the initial energy of the atom energy is equal to the energy of the emitted photon.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{photon}}={\mathrm{E}}_2-{\mathrm{E}}_1 \\ =\left(-0.85\mathrm{eV}\right)-\left(-3.4\mathrm{eV}\right) \\ =2.55\mathrm{eV} \end{gathered}$$

Therefore, the energy of the photon emitted is 2.55.

(b) Define the higher quantum number:

Rearrange the equation (1) as below.

$n_1=\sqrt{-\frac{13.6eV}{E_1}}$

Substitute -0.85 eV for $E_1$in the above equation.

$$\begin{gathered} n_1=\sqrt{-\frac{13.6eV}{-0.85eV}} \\ =4 \end{gathered}$$

Therefore, the higher quantum number is 4.

(c) Calculate the higher quantum number:

Rearrange the equation (1).

$n_1=\sqrt{-\frac{13.6eV}{E_1}}$

Substitute -3.4 eV for ${\mathrm{E}}_1$in the above equation.

$$\begin{gathered} n_2=\sqrt{-\frac{13.6eV}{-3.4eV}} \\ =2 \end{gathered}$$

Hence, the lower quantum number is 2 .