Chapter 27.1, Problem 11PP

To determine

To Find: Maximum kinetic energy of the photoelectrons.

Answer to Problem 11PP

  $5\cdot 12\times {10}^{-19}\text{ J}$

Explanation of Solution

Given data:

The stopping potential of photocell is $3.2\text{ V}$ .

Formula used:

Stopping potential for an 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 $=1\cdot 6\times {10}^{-19}\text{ C}$

  $\begin{array}{l}K{E}_{\max }=1\cdot 6\times {10}^{-19}\times 3\cdot 2\\ K{E}_{\max }=5\cdot 12\times {10}^{-19}\text{ J}\end{array}$

Conclusion:

The kinetic energy of ejected electrons is $5\cdot 12\times {10}^{-19}\text{ J}$ .