Chapter 17, Problem 8P

To determine

The potential difference needed to give a helium nucleus $65.0\text{keV}$ of kinetic energy.

Answer to Problem 8P

The potential difference is $32.5\times {10}^3\text{V}$ .

Explanation of Solution

Given info:

The charge of helium is $q=2e$ .

Kinetic energy is $KE=65.0\text{keV}$ .

Formula used:

Kinetic energy in accelerating the electron through the potential difference is,

  $KE=q\Delta V$

Calculation:

Substituting the given values, we get

  $\begin{array}{c}65.0\text{keV}=\left(2e\right)\left(\Delta V\right)\\ \Delta V=\frac{65.0\times {10}^3\text{eV}}{2e}\\ =32.5\times {10}^3\text{V}\end{array}$

Conclusion:

Thus, the potential difference is $32.5\times {10}^3\text{V}$ .