Chapter 27.2, Problem 27PP

To determine

To Calculate: Potential difference to accelerate an electron.

Answer to Problem 27PP

  $96.32\text{ V}$

Explanation of Solution

Given data:

Wavelength of electron, $\lambda =0.125\text{nm}$

Formula used:

de Broglie wavelength of electron is related to its velocity:

  $\lambda =\frac{h}{mv}$

Here, h is the Planck’s constant, m is the mass of electron and v is the velocity of the electron.

The potential difference V can be obtained by:

  $\begin{array}{l}eV=\frac{1}{2}m{v}^2\\ h=6\cdot 6\times {10}^{-34}\text{J}\cdot \text{s}\\ m=9\cdot 1\times {10}^{-31}\text{kg}\\ e=1\cdot 6\times {10}^{-19}\text{C}\end{array}$

Calculation:

The velocity of the electron can be obtained using the de Broglie wavelength.

  $\begin{array}{l}\lambda =\frac{h}{mv}\\ v=\frac{h}{m\lambda }\end{array}$

  $\begin{array}{l}eV=\frac{1}{2}m{v}^2\\ eV=\frac{1}{2}m{\left(\frac{h}{m\lambda }\right)}^2\\ V=\frac{h^2}{2me{\lambda }^2}\end{array}$

  $\begin{array}{l}V=\frac{{\left(6\cdot 62\times {10}^{-34}\right)}^2}{2\times 9\cdot 1\times {10}^{-31}\times 1\cdot 6\times {10}^{-19}\times {\left(0\cdot 125\times {10}^{-9}\right)}^2}\\ V=96.32\text{ V}\end{array}$

Conclusion:

The potential difference to accelerate an electron is $96.32\text{ V}$ .