JEE Main 2026 — 28 January, Evening Shift — Question 65

When a charged particle is accelerated through a potential difference $V$, it gains kinetic energy equal to electrical work done.

$$K.E. = qV$$

Now kinetic energy is also related to momentum as

$$K.E. = \frac{p^2}{2m}$$

Equating both expressions, we have $\frac{p^2}{2m} = qV$

So momentum becomes, $p = \sqrt{2mqV}$

The de Broglie wavelength is $\lambda = \frac{h}{p}$

Substituting momentum, we have $\lambda = \frac{h}{\sqrt{2mqV}}$

Since $q$ and $V$ are same for both particles, thus we have $\lambda \propto \frac{1}{\sqrt{m}}$

Taking ratio, we have

= \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{4}{1}} = 2$$