Chapter 27, Problem 58A

(a)

To determine

To Calculate: Potential difference experienced by electron.

Answer to Problem 58A

  $46.6\text{V}$

Explanation of Solution

Given data:

de Broglie wavelength of the electron, $\lambda =0.18\text{nm}$

Formula used:

de Broglie equation:

  $\begin{array}{l}\lambda =\frac{h}{p}\text{ where }p=\sqrt{2m_{e}KE}\\ KE=eV\\ \lambda =\frac{2h}{\sqrt{2m_{e}eV}}\\ \Rightarrow {\lambda }^2=\frac{h^2}{2m_{e}eV}\\ V=\frac{h^2}{2m_{e}e{\lambda }^2}\end{array}$

Calculation:

Substitute the values and solve:

  $\begin{array}{l}V=\frac{h^2}{2me{\lambda }^2}\\ V=\frac{{\left(6.63\times {10}^{-34}\right)}^2}{\left(2\right)\left(9.1\times {10}^{-31}\right)\left(1.6\times {10}^{-19}\right){\left(0.18\times {10}^{-9}\right)}^2}\\ V=46.6\text{V}\end{array}$

Conclusion:

Potential difference experienced by electron is $46.6\text{V}$ .

(b)

To determine

To Calculate: Potential difference experienced by a proton.

Answer to Problem 58A

  $\text{0}\text{.025 V}$

Explanation of Solution

Given data:

de Broglie wavelength of the proton, $\lambda =0.18\text{nm}$

Formula used:

de Broglie equation:

  $\begin{array}{l}\lambda =\frac{h}{p}\text{ where }p=\sqrt{2m_{p}KE}\\ KE=eV\\ \lambda =\frac{2h}{\sqrt{2m_{p}eV}}\\ \Rightarrow {\lambda }^2=\frac{h^2}{2m_{p}eV}\\ V=\frac{h^2}{2m_{p}e{\lambda }^2}\end{array}$

Calculation:

Substitute the values and solve:

  $\begin{array}{l}V=\frac{h^2}{2m_{p}e{\lambda }^2}\\ V=\frac{{\left(6.63\times {10}^{-34}\right)}^2}{\left(2\right)\left(1.67\times {10}^{-27}\right)\left(1.6\times {10}^{-19}\right){\left(0.18\times {10}^{-9}\right)}^2}\\ V=253.87\times {10}^{-4}\text{V = 0}\text{.025 V}\end{array}$

Conclusion:

Potential difference experienced by proton is $\text{0}\text{.025 V}$ .