Chapter 27, Problem 46P

To determine

The voltage required to produce the electron wavelengths of $0.26\text{nm}$.

Answer to Problem 46P

Solution:

The voltage required is 22.2 V.:

Explanation of Solution

Given:

Wavelengths of electron: $0.26\text{nm}$

Formula Used:

$\begin{array}{l}KE=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}\\ \end{array}$

Calculation:

By using the formula, $\text{KE}=\frac{h^2}{2m{\lambda }^2}$

Kinetic energy can be calculated as

$\begin{array}{l}\text{KE}=\frac{h^2}{2m{\lambda }^2}\\ KE=\frac{{(6.63\times {10}^{-34}\text{J}\text{s})}^2}{2(9.11\times {10}^{-31}\text{kg}){(0.26\times {10}^{-9}\text{m})}^2}=3.57\times {10}^{-18}\text{J}\end{array}$

Now on converting Joule to electron-volt 22

$\frac{3.57\times {10}^{-18}\text{J}}{1.6\times {10}^{-19}}=22.2\text{eV}$

So, 22.2 V is required to produce the electron wavelengths of $0.26\text{nm}$.