An `alpha`-particle and a proton are accelerated from rest by the same potential, thenthe ratio of their de-Broglie wavelength is

Options

A. `2 sqrt2 : 1`
B. `1 : 2 sqrt2`
C. `1 : 2`
D. `2 : 1`

Answer

B

Explanation

de-Broglie.s wavelength,
`lamda = (h)/(p) = (h)/(sqrt(2mE)) = (h)/(sqrt(2meV))`
The particles are at same potential, so
`(lamda_(alpha))/(lamda_(p)) = (h//sqrt(2m_(alpha)e_(alpha)V))/(h//sqrt(2m_(p) e_(p)V))`
Mass of `alpha`-particle =4 times mass of proton.
Charge of `alpha`-particle =2 times charge of proton.
So, `(lamda_(alpha))/(lamda_(p)) = sqrt((m_(p) e_(p))/(4m_(p) 2e_(p))) = (1)/(2sqrt2)`