Chapter 27, Problem 55A

(a)

To determine

To Find: The velocity of an electron.

Answer to Problem 55A

  $4.19\times {10}^7\text{m/s}$

Explanation of Solution

Given data:

Potential difference, $V=5.0\times {10}^3\text{V}$

Formula used:

The potential difference gives rise to kinetic energy which raises velocity of electron from $0\text{ to }v\text{ m/s}$ .

  $\begin{array}{l}eV=\frac{1}{2}m{v}^2-0\\ v=\sqrt{\frac{2eV}{m}}\end{array}$

Here, e is the charge of electron, V is the potential difference and m is the mass of the electron.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}v=\sqrt{\frac{2eV}{m}}\\ v=\sqrt{\frac{2\times 1.6\times {10}^{-19}\text{C}\times 5.0\times {10}^3\text{V}}{9.1\times {10}^{-31}\text{kg}}}\\ v=4.19\times {10}^7\text{m/s}\end{array}$

Conclusion:

The velocity of the electron is $4.19\times {10}^7\text{m/s}$ .

(b)

To determine

To Find: The wavelength of the electron.

Answer to Problem 55A

  $0.17\stackrel{\text{o}}{\text{A}}$

Explanation of Solution

Given:

The velocity of the electron is $4.19\times {10}^7\text{m/s}$ .

Formula used:

de Broglie’s equation is:

  $\lambda =\frac{h}{mv}$

  $\lambda =$ de Broglie wavelength

  $h=$ Planck’s constant

m = mass

v = velocity

Calculation:

Now using de Broglie’s equation:

  $\lambda =\frac{h}{mv}$

  $\lambda =\frac{6.63\times {10}^{-34}}{9.1\times {10}^{-31}\times 4.19\times {10}^7}$

  $\begin{array}{l}\lambda =0.17\times {10}^{-10}\text{m}\\ \lambda =0.17\stackrel{\text{o}}{\text{A}}\end{array}$

Conclusion:

Electron’s wavelength is $0.17\stackrel{\text{o}}{\text{A}}$ .