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

Answer

C

Explanation

When a charged particle of charge q , mass m enters perpendicularly to the magnetic induction B of a magnetic field , it will experience a magnetic force
`F = qvB sin 90^(@) = qvB`
that provides a centripetal acceleration `v^2//r`.
So, `qvB = (mv^2)/r" or " mv= qBr`
The de Broglic wavelength
`lamda = h/(mv) = h/(qBr), lamda_(alpha) / lamda_(p) = (q_(p)r_(p))/(q_alpha r_alpha)`
Since , `(r_alpha)/(r_p) =1 and (q_(alpha))/(q_p)=2 :. lamda_(alpha)/lamda_p = 1/2`