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 such that radius of curvature of their path is same. The ratio of their de-Broglie wavelengths is _______

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Detailed Solution

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

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

$\begin{array}{l} In our case, F_{B} = q v B [As θ = 90^{\circ}] \\ Hence B q v = \frac{m v^{2}}{r} \Rightarrow m v = q B r \end{array}$

$\begin{array}{l} The de-Broglie wavelength λ = \frac{h}{m v} = \frac{h}{q B r} \Rightarrow \frac{λ_{α - particle}}{λ_{proton}} = \frac{q_{p} r_{p}}{q_{α} r_{α}} \\ Since, \frac{r_{α}}{r_{p}} = 1 and \frac{q_{α}}{q_{p}} = 2, \Rightarrow \frac{λ_{α}}{λ_{p}} = \frac{1}{2} \end{array}$