The radius of an  $\alpha -$ particle moving in a circle in a constant magnetic field is half the radius of an electron moving in circular path in the same field. The de-Broglie wavelength of $\alpha -$ particle  is n times that of the electron. Find n ( an integer)____

Radius of a charged particle in a magnetic field is given by,

$\mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}$

$\therefore \text{ momentum }=\mathrm{p}=\mathrm{mV}=\mathrm{RqB}$

It is given that radius of alpha particle is half of electron. We know that the charge of an alpha particle is twice of that of an electron. 

$\frac{p_{\alpha }}{p_e}=\frac{R_{\alpha }{q}_{\alpha }}{R_e{q}_e}=\frac{1}{2}\times 2=1$

Therefore, momentum remains same for both electron and alpha particle.

The de-Broglie wavelength, $\lambda =\frac{h}{p}$

${\lambda }_{\alpha }=\frac{p_e}{p_{\alpha }}{\lambda }_e={\lambda }_e$