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Question:

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50V. Another particle B of mass '4m' and charge 'q' is accelerated by a potential difference of 2500V. The ratio of de-Broglie wavelengths λA/λB is close to:

0.07

14.14

4.47

10.00

Solution:

K.E acquired by charge = K = qV
λ = h/p = h/√(2mK) = h/√(2mqV)
λA/λB = √(2mB q VB) / √(2mA q VA) = √(4m × 2500) / √(m × 50) = √(10000m/50m) = √200 = 14.14