An `alpha`-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de Broglie wavelengths associated with them.

Explanation

The kinetic energy of a particle of mass m,
`E = (p^(2))/(2m)`, where `p = sqrt(2mE)`
So, de Broglie wavelength, `lamda = (h)/(p) = (h)/(sqrt(2mE))`
Now, if a particle of charge q is accelerated by applying potential difference V, then
`E = qV or, lamda = (h)/(sqrt(2mqV))`
`:. lamda prop (1)/(sqrtmq)` when V is constant
Hence, `(lamda_(1))/(lamda_(2)) = sqrt((m_(2).q_(2))/(m_(1).q_(1)))`
For proton and `alpha-`particle, `(m_(alpha))/(m_(p)) = 4, (q_(alpha))/(q_(p)) = 2`
`:. (lamda_(p))/(lamda_(alpha)) = sqrt((m_(alpha).q_(alpha))/(m_(p).q_(p))) = sqrt(4 xx 2) = 2 sqrt2`
Hence, the required ratio is `2 sqrt2:1`