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
$F=q v B \sin 90^{\circ}=q v B$
that provides a centripetal acceleration $v^{2} / r$.
So, $q v B=\frac{m v^{2}}{r}$ or $m v=q B r$
The de Broglie wavelength
$\lambda=\frac{h}{m v}=\frac{h}{q B r} $;
$ \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 $
$\therefore \frac{\lambda_{\alpha}}{\lambda_{p}}=\frac{1}{2}$