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`

Explanation

When a charged particle of charge `q`, mass `m` enters perpendicularly to the magnetic induction `vecB` of a magnetic field, it will experience a magnetic force `F=q(vecvxxvecB)=q vB sin 90^(@)=qvB` that provide a centripetal acceleration `(v^(2))/(r )`
`rArr qvB=(mv^(2))/(r )rArrmv=qBr`
`rArr` The `de`-Broglie wavelength `lambda=(h)/(mv)=(h)/(qBr)`
`rArr (lambda_(alpha-"particle"))/(lambda_("proton"))=(q_(p)r_(p))/(q_(alpha)r_(alpha))`
Since `(r_(alpha))/(r_(p))=1` and `(q_(alpha))/(q_(p))=2`
`rArr (lambda_(alpha))/(lambda_(p))=1//2`