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An $α$ -particle and a proton are fired through the same magnetic fields which is perpendicular to their velocity vectors. The $α$ -particle and the proton move in such a way that the radius of curvature of their path is same. Find the radius of their de Broglie wavelengths.

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$\frac{λ_{α}}{λ_{p}} = \frac{2}{5}$

$\frac{λ_{p}}{λ_{a}} = \frac{9}{4}$

$\frac{λ_{α}}{λ_{p}} = \frac{1}{2}$

$\frac{λ_{p}}{λ_{a}} = \frac{3}{2}$

answer is C.

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Detailed Solution

Magnetic force experienced by a charged particle in a magnetic field is given by,

$F_{B} = q \vec{v} \times \vec{B} = q v B \sin θ$

In our case $F_{B} = q v B$ [as $θ = 90 °$ ]

Hence, $B q v = \frac{m v^{2}}{r}$ or $m v = q B r$

The de Broglie wavelength

$λ = \frac{h}{m v} = \frac{h}{q B r} \Rightarrow \frac{λ_{α}}{λ_{p}} = \frac{q_{p} r_{p}}{q_{α} r_{α}}$

Since $\frac{r_{α}}{r_{p}} = 1$ and $\frac{q_{α}}{q_{p}} = 2$ ,

$∴ \frac{λ_{α}}{λ_{p}} = \frac{1}{2}$

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