Search any question & find its solution
Question:
Answered & Verified by Expert
An electron accelerated through a potential difference ' $V_1$ ' has a de-Broglie wavelength ' $\lambda$ '. When the potential is changed to ' $\mathrm{V}_2$ ' its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)$ is
Options:
Solution:
2918 Upvotes
Verified Answer
The correct answer is:
9:4
For electron, de Broglie wavelength, $\lambda=\frac{1.228}{\sqrt{\mathrm{V}}}$ Given: $\lambda_2=\lambda_1+0.5 \lambda_1=1.5 \lambda_1$
$\begin{aligned}
& \therefore \quad \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \\
& \quad \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}
\end{aligned}$
$\begin{aligned}
& \therefore \quad \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \\
& \quad \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.