Chapter 27, Problem 83GP

(a)

To determine

To draw: The energy level diagram of the given element.

Explanation of Solution

Given:

Ionization energy is $11.5\text{ eV}$

Absorption spectrum with lines at $\text{2}\text{.5 eV; 4}\text{.7 eV; 5}\text{.1 eV}$

The energy level diagram is shown below.

  

According to the above energy level diagram, the ground state level is represented as $E_1$ similarly all the energy levels are represented as $E_2;E_3;E_4$

First transition takes place from $E_4\text{ to }{E}_3$

Second transition takes place from $E_4\text{ to }{E}_2$

Third transition will be from $E_4\text{ to }{E}_1$

Conclusion:

Hence, the transition energy from $E_4$ to the ground state is 5.1 eV.

(b)

To determine

The state in which the atom was before absorbing the photon.

Energies of the photons that can subsequently be emitted by the atom.

Answer to Problem 83GP

An atom is in ground state.

The energies are: $\Delta E_{\text{I}}=0.4\text{ eV,}\Delta E_{\text{II}}=2.6\text{ eV, }\Delta E_{\text{III}}=5.1\text{ eV}$

Explanation of Solution

Given:

Ionization energy is $11.5\text{ eV}$

  $E_1=0;E_2=\text{2}\text{.5 eV;}{E}_3=\text{4}\text{.7 eV;}{E}_4=\text{5}\text{.1 eV}$

Calculation:

The atom was in the ground state before absorbing the photon, because the transition from $E_4$ to $E_1$ is 5.1 eV.

Now, the energies of the photons emitted by the atom can be calculated as follows:

Now, the first transition energy is given as:

  $\begin{array}{l}\Delta E_{\text{I}}=E_4-E_3=\text{5}\text{.1 eV}-\text{4}\text{.7 eV}\\ \Delta E_{\text{I}}=0.4\text{ eV}\end{array}$

The second transition energy:

  $\begin{array}{l}\Delta E_{\text{II}}=E_4-E_2=\text{5}\text{.1 eV}-2.5\text{ eV}\\ \Delta E_{\text{II}}=2.6\text{ eV}\end{array}$

The third transition energy:

  $\begin{array}{l}\Delta E_{\text{III}}=E_4-E_1=\text{5}\text{.1 eV}-0\text{ eV}\\ \Delta E_{\text{III}}=5.1\text{ eV}\end{array}$

Conclusion:

The atom is in ground state before absorbing the photon. The energies are: $\Delta E_{\text{I}}=0.4\text{ eV,}\Delta E_{\text{II}}=2.6\text{ eV,}\Delta E_{\text{III}}=5.1\text{ eV}$