A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mas '4m' and charge 'q' is accelerated by a potential difference of 2500V. The ratio of de-Broglie wavelength $\frac{λ_A}{λ_B}$ is

14.14

$\text{Debroglie wavelength }\lambda = \frac{h}{p} = \frac{h}{\sqrt {2mE}} = \frac{h}{\sqrt {2mqV}} $

$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{m_B}{m_A}} \sqrt{\frac{V_B}{V_A}} = \sqrt{\frac{4m}{m}} \sqrt{\frac{2500}{50}} = 10\sqrt 2 =14.14 $