Question

An electron is accelerated from rest through a potential difference of \(\mathrm{V}\) volt. If the de Broglie wavelength of the electron is \(1.227 \times 10^{-2} n m\), the potential difference is :
A \(10 \mathrm{~V}\)
в \(10^{2} \mathrm{v}\)
c \(10^{3} \mathrm{v}\)
(D) \(10^{4} \mathrm{v}\)

Answer 1

Solution:
$$
\begin{aligned}
&\lambda=1.227 \times 10^{-2} m \\
&=0.1227 \AA \\
&\lambda=\frac{12.27}{\sqrt{V}} \AA \\
&0.1227=\frac{12.27}{\sqrt{V}} \AA \\
&\sqrt{V} \times 10^{2} \\
&\Rightarrow V=10^{4}
\end{aligned}
$$

Answer 2

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Answer 3

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