Chapter 27.2, Problem 31SSC

To determine

To Calculate: The de Broglie wavelength of electron.

Answer to Problem 31SSC

  $\text{1}\text{.09}\stackrel{\text{o}}{\text{A}}$

Explanation of Solution

Given data:

Potential difference is $V=125\text{V}$ .

Formula used:

de Broglie wavelength $(\lambda )$ can be obtained by:

  $\lambda =\frac{h}{mv}=\frac{h}{p}$

  $p$ can be written as

  $p=\sqrt{2mKE}$

Kinetic energy in terms of potential difference is

  $KE=eV$

Therefore, de Broglie wavelength in terms of potential difference is

  $\lambda =\frac{h}{\sqrt{2meV}}$

Where,

  $\lambda =$ de Broglie wavelength

  $h=$ Planck’s constant

  $m=$ mass of electron

v = velocity of electron

  $p=$ momentum

  $KE=$ Kinetic Energy

  $e=$ Charge of electron

  $V=$ Potential difference

Calculation:

Planck’s constant, $h=6.6\times {10}^{-34}\text{J}\cdot {\text{s}}^{\text{-1}}$

Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

Charge of electron, $e=1.6\times {10}^{-19}\text{C}$

Given Potential difference, $V=125\text{V}$

Substitute the values and solve:

  $\lambda =\frac{h}{\sqrt{2meV}}$

  $\begin{array}{l}\lambda \text{ =}\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 125}}\\ \lambda =1.09\times {10}^{-10}\text{m}\text{=1}\text{.09}\stackrel{\text{o}}{\text{A}}\end{array}$

Conclusion:

Therefore, de Broglie wavelength is $\text{1}\text{.09}\stackrel{\text{o}}{\text{A}}$ .