Chapter 28.1, Problem 6PP

(a)

To determine

To Find: The energy of the photon emitted by the mercury atom.

Answer to Problem 6PP

  $\text{2}\text{.15}\text{eV}$

Explanation of Solution

Given data:

For a particular transition, the energy of a mercury atom drops from $8.82\text{ eV}$ to $6.67\text{ eV}$ .

  $\begin{array}{l}{E}_2=8.82\text{eV}\\ E_1=6.67\text{eV}\end{array}$

Formula Used:

  $\Delta E=E_2-E_1$

  $$

Calculation:

  $\begin{array}{l}\Delta E=E_2-E_1\\ \Delta E=8.82\text{eV}-6.67\text{eV}\\ \Delta E\text{=2}\text{.15}\text{eV}\end{array}$

Conclusion:

The energy of the photon emitted by the mercury atom is $\text{2}\text{.15}\text{eV}$ .

(b)

To determine

To Find: The wavelength of the photon emitted by the mercury atom.

Answer to Problem 6PP

  $576.7\text{nm}$

Explanation of Solution

Given data:

The energy of the photon emitted by the mercury atom is $\text{2}\text{.15}\text{eV}$ .

Formula used:

  $\Delta E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }$

Here, $\lambda$ is the wavelength of the photon.

Calculation:

  $\begin{array}{l}\Delta E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }\\ 2.15\text{eV}=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }\\ \lambda =576.7\text{nm}\end{array}$

Conclusion:

The wavelength of the photon emitted is $576.7\text{nm}$ .