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
Concept. $\lambda \propto \dfrac{1}{\sqrt V}$.
Step 1. Ratio. $\dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2 = 1.5^2 = \dfrac94$.
Step 2. Compare. $\alpha = 4$.
Answer. $4$.
Solution by Sreeraj P, M.Sc Physics
