Chapter 27, Problem 44P

To determine

The de Broglie wavelength of an electron.

Answer to Problem 44P

Solution:

The de Broglie wavelength of an electron $6.5\times {10}^{-12}\text{m}$.

Explanation of Solution

Given:

The electron is accelerated by a potential of: $35,000\text{V}$

Formula Used:

1. $v=\frac{h}{m\lambda }$

2. $\frac{1}{2}m{v}^2=eV$

Where, v is the velocity of electron

h is Plank’s constant

m is mass of electron

e is the charge on electron

$\lambda$ is the wavelength of the electron.

Calculation:

In the formula 2., we will multiply both the sides by 2m

$\begin{array}{l}2m.\frac{1}{2}m{v}^2=2m.eV\\ m^2{v}^2=2m.eV\\ mv=\sqrt{2m.eV}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_(1)\end{array}$

Now, from the formula (1)$v=\frac{h}{m\lambda }$, the value of $mv$ can be given as

$v=\frac{h}{m\lambda }$

$mv=\frac{h}{\lambda }\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_(2)$

Now, equating equation (1)and (2), we get

$\begin{array}{l}\frac{h}{\lambda }=\sqrt{2meV}\\ or,\lambda =\frac{h}{\sqrt{2meV}}\\ \lambda =\frac{6.6\times {10}^{-34}\text{J}\text{.s}}{\sqrt{\text{2×9}{\text{.11×10}}^{\text{-31}}\text{kg×1}{\text{.6×10}}^{\text{-19}}\text{C×35,000 V}}}=6.5\times {10}^{-12}\text{m}\end{array}$

So, the electron accelerated from rest by a potential of 35,000 V has a wavelength of $6.5\times {10}^{-12}\text{m}$. This wavelength is very much small than the size of the neck of the tube which is 5 cm so there will be no problem of blurring the picture on the CRT because of diffraction.