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The ratio of de Broglie wavelength of an $α$ 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 B is perpendicular to the velocity vectors of the $α$ -particle and the proton is

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$\frac{1}{2}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

When a charged particle (charge q, mass m) enters perpendicularly in a magnetic field (B) then, radius of the path described by it

$r = \frac{mv}{qB}$

$\Rightarrow mv = qBr$

Also de-broglie wavelength $λ = \frac{h}{mv}$

$\Rightarrow λ = \frac{h}{qBr} \Rightarrow \frac{λ_{α}}{λ_{p}} = \frac{q_{p} r_{p}}{q_{a} r_{a}} = \frac{1}{2}$

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