Chapter 28, Problem 32A

a.

To determine

To calculate: Energy needed to ionize the atom.

Answer to Problem 32A

Energy needed to ionize the atom = $\underset{\_}{2.72\text{ eV}}$ .

Explanation of Solution

Given:

The mercury atom is in an excited state at $E_6$ energy level.

Formula used:

Energy change of an atom is given by,

  $\Delta E=E_f-E_i$

Where $E_f$ is the final energy and $E_i$ is the initial energy.

Calculation:

According to the given diagram,

  $\begin{array}{l}{E}_{\text{ionization}}=10.44\text{ eV}\\ E_6=7.72\text{ eV}\end{array}$

Energy needed to ionize the atom is calculated as:

  $\begin{array}{l}\Delta E=E_f-E_i\\ \text{ =}10.44-7.72\\ =2.72\text{ eV}\end{array}$

Conclusion:

Energy needed to ionize the atom = $\underset{\_}{2.72\text{ eV}}$

b.

To determine

To calculate: Energy released if the atom is dropped to $E_2$ energy level.

Answer to Problem 32A

Energy released if the atom is dropped to $E_2$ energy level = $\underset{\_}{3.06\text{ eV}}$ .

Explanation of Solution

Given:

The mercury atom is in an excited state at $E_6$ energy level.

Formula used:

Energy released by an atom is given by,

  $\Delta E=E_i-E_f$

Where $E_i$ is the initial energy and $E_f$ is the final energy and $E_i$ is the initial energy.

Calculation:

According to the given diagram,

  $\begin{array}{l}{E}_6=7.72\text{ eV}\\ E_2=4.66\text{ eV}\end{array}$

Energy released if the atom is dropped to $E_2$ energy level is calculated as:

  $\begin{array}{l}\Delta E=E_i-E_f\\ \text{ =}7.72-4.66\\ =3.06\text{ eV}\end{array}$

Conclusion:

Energy released if the atom is dropped to $E_2$ energy level = $\underset{\_}{3.06\text{ eV}}$ .