Chapter 27.1, Problem 9PP

To determine

To Find: Kinetic energy given by incident light to electron.

Answer to Problem 9PP

  $K{E}_{\max }=8\cdot 16\times {10}^{-19}\text{ J}$

Explanation of Solution

Given data: Stopping potential for photoelectric cell, $V_o=5.1\text{V}$

Formula used:

Stopping potential for a electric cell is related to kinetic energy of the electrons ejected.

  $K{E}_{\max }=e{V}_0$

  $\therefore K{E}_{\max }=$ kinetic energy of electron when it leaves the metal surface.

  $e{V}_0\to$ potential energy of electron

Calculation:

  $e=$ change of electron $=1\cdot 6\times {10}^{-19}\text{ C}$

Substitute the values and solve:

  $\begin{array}{l}K{E}_{\max }=1\cdot 6\times {10}^{-19}\times 5\cdot 1\text{J}\\ K{E}_{\max }=8\cdot 16\times {10}^{-19}\text{ J}\end{array}$

Conclusion:

The kinetic energy of ejected electrons is $8\cdot 16\times {10}^{-19}\text{ J}$ .