A proton and an `alpha` particle are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths is

Options

A. `sqrt8`
B. `(1)/(sqrt8)`
C. 1
D. 2

Answer

A

Explanation

When a charged particle of charge q and mass m is accelerated under a potential difference V , let v be velocity acquired by the particle , Then
`q V = (1)/(2) mv^(2) or mv = sqrt(2 mq V)`
de Broglie wavelength , `lambda = (h)/(mv) = (h)/(sqrt(2 mqV))`
`lambda prop (1)/(sqrt(mq))` for the same value of V .
`therefore (lambda_(p))/(lambda_(alpha)) = (sqrt(m_(alpha) q_(alpha)))/(sqrt(m_(p) q_(p))) = sqrt((4m)/(m) xx (2e)/(e)) = sqrt8`