Q 12-11-080JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases by $50\%$. The value of $\left(\dfrac{V_1}{V_2}\right)$ is equal to
Answer: (B) $\dfrac94$
$\lambda\propto\dfrac1{\sqrt V}$. With $\lambda_2=1.5\lambda_1$: $\dfrac{V_1}{V_2}=\left(\dfrac{\lambda_2}{\lambda_1}\right)^2=\dfrac94$.
Solution by Sreeraj P, M.Sc Physics