If the two particles of the same mass and same charge are accelerated with a potential difference of 120 V and 480 V respectively, the ratio of de Broglie's wavelengths associated with the two particles respectively is

2 : 1

The correct answer is Option (2) → 2 : 1

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

For charge $q$, mass $m$, accelerated through potential $V$:

$\frac{1}{2} m v^2 = qV \;\;\Rightarrow\;\; p = \sqrt{2mqV}$

So, $\lambda = \frac{h}{\sqrt{2mqV}} \;\;\propto\;\; \frac{1}{\sqrt{V}}$

Given: $V_1 = 120 \, \text{V}$, $V_2 = 480 \, \text{V}$

Ratio: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{480}{120}} = \sqrt{4} = 2$

Answer: $\lambda_1 : \lambda_2 = 2 : 1$