Chapter 28, Problem 3STP

To determine

To Identify:The wavelength of light.

Answer to Problem 3STP

The answer is D- $1020\text{ nm}$ .

Explanation of Solution

Given data:

The energy levels for a mercury atom are given in the diagram.

Formula used:

  $\Delta E=\frac{hc}{\lambda }$

Here, h is Planck’s constant and c is the speed of light.

  $\text{ [}\lambda \text{=405 nm, c=3}\times {\text{10}}^8\text{m/s]}$

Calculation:

  $\begin{array}{l}\Delta E=\frac{hc}{\lambda }\\ \lambda =\frac{1240\text{ eV}\text{.nm}}{1.22}\\ \lambda =1020\text{ nm}\end{array}$

Conclusion:

Therefore, answer is D- $1020\text{ nm}$ .