An `alpha-"particle"` is accelerated through a potential difference of `V` volts from rest. The de-Broglie's wavelengths associated with it is.

Options

A. `sqrt((150)/(V)) Å`
B. `(0.286)/(sqrt(V)) Å`
C. `(0.101)/(sqrt(V)) Å`
D. `(0.983)/(sqrt(V)) Å`

Answer

C

Explanation

( c) For an a particle , `lamda = (0.101)/(sqrt(V)) Å`.