The ratio of the de-Broglie wavelengths associated with protons accelerated through a potential of 128 V and alpha particles accelerated through a potential of 64 V will be

2 : 1

The correct answer is Option (3) → 2 : 1

For a particle accelerated through potential V:

$\lambda = \frac{h}{\sqrt{2m e V}}$

Let $\lambda_1$ be for proton and $\lambda_2$ for alpha particle.

$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 V_2}{m_1 V_1}}$

Given:
$m_1 = m_p$ , $m_2 = 4m_p$ , $V_1 = 128$ V , $V_2 = 64$ V

$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{4m_p \times 64}{m_p \times 128}} = \sqrt{\frac{256}{128}} = \sqrt{2}$

∴ Ratio = $\lambda_1 : \lambda_2 = \sqrt{2} : 1$