Question

A proton and α-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelength will be(a) 1 : 1(b) 1 : 2(c) 2 : 1(d) 2√2 : 1

Answer 1

answer is option D

wavelength =

$h \div \sqrt{2m{qv} }$

where v is the applied potential difference.

as charge of alpha is +2e and it's mass is 4amu.

and for proton it is +1e and 1 amu .

that is √8:1