Question

The ratio of de Broglie wavelength of $\alpha$ -particle to that of a proton being subjected to the same magnetic field so that the radii of their paths are equal to each other assuming the field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$ -particle and the proton is

• A. 1
• B. $1/4$
• C. $1/2$
• D. 2

Answer 1

Answer: (C)

When a charged particle of charge $q$, mass $m$ enters perpendicularly to the magnetic induction $B$ of a magnetic field, it will experience a magnetic force
$F=qvB\sin 90^{\circ }=qvB$
that provides a centripetal acceleration $v^2/r$.
So, $qvB=\frac{m{v}^2}{r}$ or $mv=qBr$
The de Broglie wavelength
$\lambda =\frac{h}{mv}=\frac{h}{qBr}$;
$\frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{q_p{r}_p}{q_{\alpha }{r}_{\alpha }}$
Since, $\frac{r_{\alpha }}{r_p}=1$ and $\frac{q_{\alpha }}{q_p}=2$
$\therefore \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{1}{2}$