Question

An electron is accelerated from rest through a potential difference of $Vvolt$. If the de Broglie wavelength of the electron is $1.227\times {10}^{-2}nm$, the potential difference is

Answer 1

Step 1: Given data

1. The potential difference of the electron is $Vvolt.$
2. The de Broglie wavelength of the electron is $\lambda =1.227\times {10}^{-2}nm.$

Step 2: De Broglie wavelength

1. The De Broglie wavelength of a particle of mass $m$ and velocity $v$ is defined by the form, $\lambda =\frac{h}{mv}$, where, $h$ is the Plank constant.
2. Another form of de Broglie wave is $\lambda =\frac{1.227}{\sqrt{V}}nm$, where, V is the voltage of the electron.

Step 3: Finding the potential difference

As we know, the de Broglie wavelength is, $\lambda =\frac{1.227}{\sqrt{V}}nm$.

So,

$$\begin{gathered} V={\left(\frac{1.227}{\lambda }\right)}^2={\left(\frac{1.227}{1.227\times {10}^{-2}}\right)}^2 \\ orV={10}^{4}volt. \end{gathered}$$

Therefore, the potential difference is ${10}^{4}volt.$