Chapter 27, Problem 42A

To determine

To Find: The potential difference needed to stop the electrons.

Answer to Problem 42A

  $3\text{ V}$

Explanation of Solution

Given data:

  $K.E_{\max }=\text{4}\text{.8}\times {\text{10}}^{-19}\text{J}$

Formula used:

The maximum kinetic energy and potential difference are related as follows:

  $K.E_{\max }=eV$

Here, e is the electronic charge.

Calculation:

  $\begin{array}{l}\text{K}\text{.E}=\text{4}\text{.8}\times {\text{10}}^{-19}\\ ev=\text{4}\text{.8}\times {\text{10}}^{-19}\\ 1.6\times {10}^{-19}v=\text{4}\text{.8}\times {\text{10}}^{-19}\text{ powers cancel out}\\ v=3\text{ volt}\end{array}$

Conclusion:

Thus, the potential difference needed to stop the electrons is 3 V.