Q 12-11-036JEE MainJEE Main 2026 (6 Apr, Shift 2)Easy
The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 4
$\lambda = \dfrac{h}{\sqrt{2meV}} \propto \dfrac{1}{\sqrt{V}}$, so $\dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2 = (1.5)^2 = \dfrac{9}{4}$.
$\alpha = 4$.
Solution by Sreeraj P, M.Sc Physics