JEE Main 2026 — Dual Nature of Matter Question with Solution

The de Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda =\frac{h}{\sqrt{2meV}}$.

This implies $\lambda \propto \frac{1}{\sqrt{V}}$.

Given that the new wavelength ${\lambda }_2$ is increased by $50\%$, we have ${\lambda }_2={\lambda }_1+0.5{\lambda }_1=\frac{3}{2}{\lambda }_1$.

Taking the ratio of the wavelengths, we get $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{V_2}{V_1}}$.

Substituting ${\lambda }_2=\frac{3}{2}{\lambda }_1$, we get $\frac{2}{3}=\sqrt{\frac{V_2}{V_1}}$.

Squaring both sides yields $\frac{V_2}{V_1}=\frac{4}{9}$, which gives $\frac{V_1}{V_2}=\frac{9}{4}$.

Comparing this with the given relation $\frac{V_1}{V_2}=\frac{9}{\alpha }$, we find $\alpha =4$.

Answer: $4$