The ratio of de - Broglie wavelength of `alpha `- particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector `vec(B)` is perpendicular to the velocity vectors of the `alpha` - particle and the proton is

Options

A. `1//4`
B. `1//2`
C. `1`
D. `2`

Answer

b

Explanation

`qvB=mv^(2)//r` or `mv=qBr`
De-Broglie wavelength `lambda=h/(mv)=h/(qBr)`,
so `lambda prop 1//qr`
`:. (lambda_(alpha))/(lambda_(p))=(q_(p)r_(p))/(q_(alpha)r_(alpha))=e/(2e)xxr/r=1/2`