Chapter 27, Problem 4STP

To determine

To Choose: The correct option.

Answer to Problem 4STP

Option C

Explanation of Solution

Given data:

Energy of the incident radiation, $E=5.17\text{eV}$

Work function of the photocell, $\varphi =2.31\text{eV}$

Formula used:

  $K.E.=E-\varphi$

Here, $K.E.$ is the energy of the ejected photoelectron, E is the energy of the incident radiation and $\varphi$ is the work function of the photocell.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}K.E=5.17\text{eV}-2.31\text{eV}\\ K.E.=2.86\text{eV}\end{array}$

Conclusion:

Therefore, energy of ejected photoelectron is $\text{2}\text{.86 eV}$ . Option C is correct.