The de Broglie wavelength of a charged particle accelerated through a potential difference $\mathrm{V}$ is…
The de Broglie wavelength of a charged particle accelerated through a potential difference $\mathrm{V}$ is $\lambda$. If the potential difference is increased by $21 \%$, the de Broglie wavelength of the charged particle is
$\frac{5 \lambda}{9}$
$\frac{7 \lambda}{9}$
$\frac{9 \lambda}{11}$
$\frac{10 \lambda}{11}$
Solution
De Broglie wavelength is given by
$
\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}} \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \\
& \mathrm{V}^{\prime}=\mathrm{V}+\frac{21}{100} \mathrm{~V}=1.21 \mathrm{~V} \\
& \frac{\lambda}{\lambda^{\prime}}=\frac{1}{\sqrt{\mathrm{V}}} \times \frac{\sqrt{\mathrm{V}^{\prime}}}{1} \\
& \frac{\lambda}{\lambda^{\prime}}=\sqrt{\frac{1.21 \mathrm{~V}}{\mathrm{~V}}}=\frac{11}{10} \Rightarrow \lambda^{\prime}=\frac{10}{11} \lambda
\end{aligned}
$