The ratio between the de Broglie wavelength associated with protons, accelerated through a potential of 512 V and that of alpha particles accelerated through a potential of X volts is found to be one. Find the value of X.

Explanation

de-Broglie wavelength of accelerated charge particle
`lambda = (h)/(sqrt(2mqV))`
`lambda prop (h)/(sqrt(mq V))`
Ratio of wavelength of proton and `alpha`-particle.
`(lambda_(p))/(lambda_(alpha)) = sqrt((m_(alpha) q_(alpha) V_(alpha))/(m_(p)q_(p) V_(p))) = sqrt(((m_(alpha))/(m_(p)))((q_(alpha))/(q_(p)))((V_(alpha))/(V_(p))))`
Here, `(m_(alpha))/(m_(p)) = 4, (q_(alpha))/(q_(p)) = 2, (V_(alpha))/(V_(p)) = (X)/(512), (lambda_(p))/(lambda_(alpha)) = 1`
`1 = sqrt(4 xx 2 xx ((X)/(512))) = sqrt((X)/(64)) =(X)/(64)`
X = 64 V