Chapter 27, Problem 53A

To determine

To Find: The de Broglie wavelength of an electron.

Answer to Problem 53A

  $2.43\stackrel{\text{o}}{\text{A}}$

Explanation of Solution

Given data:

Velocity of the electron is $v=3.0\times {10}^6\text{m/s}$ .

Formula used:

de Broglie equation is:

  $\lambda =\frac{h}{mv}$

  $\lambda =$ de Broglie wavelength

  $h=$ Planck’s constant $=6.626\times {10}^{-34}\text{J}\cdot \text{s}$

m = mass

v = velocity

Calculation:

Substitute the values and solve:

  $\lambda =\frac{h}{mv}$

  $\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 3\times {10}^6}\\ \lambda =\frac{6.63\times {10}^{-9}}{27.3}\\ \lambda =0.243\times {10}^{-9}\text{m}\end{array}$

  $\begin{array}{l}\lambda =2.43\times {10}^{-10}\text{m}\\ \lambda =2.43\stackrel{\text{o}}{\text{A}}\text{ (1}\stackrel{\text{o}}{\text{A}}={10}^{-10}\text{m})\end{array}$

Conclusion:

The de Broglie wavelength of an electron moving with $3\times {10}^6\text{m/s}$ is $2.43\stackrel{\text{o}}{\text{A}}$ .