An $\alpha$-particle and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The $\alpha$-particle and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths.

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

$F_{\mathrm{B}}=q\overrightarrow{v}\times \overrightarrow{B}=qvB\sin \theta$

$\begin{array}{l}\text{ In our case, }{F}_{\mathrm{B}}=qvB \left[\mathrm{as}\theta ={90}^{\circ }\right]\\ \text{ Hence, }Bqv=\frac{m{v}^2}{r}\Rightarrow mv=qBr\end{array}$

The de Broglie wavelength,

$\lambda =\frac{h}{mv}=\frac{h}{qBr}$

$\begin{array}{l}\frac{{\lambda }_{\alpha \text{ -particle }}}{{\lambda }_{\text{proton }}}=\frac{q_p{r}_p}{q_{\alpha }{r}_{\alpha }}\\ \text{ Since }\frac{r_{\alpha }}{r_p}=1\text{ and }\frac{q_{\alpha }}{q_p}=2\\ \Rightarrow \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{1}{2}\end{array}$