Question

The ratio of de-Broglie wavelength of a $\alpha$ particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal each other assuming the field induction vector $\overrightarrow{B}$ 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

The correct option is C $1/2$

When a charged particle of charge $q$, mass $m$ enters perpendicularly to the magnetic induction $\overrightarrow{B}$ of a magnetic field, it will experience a magnetic force
$F=q(\overrightarrow{v}\times \overrightarrow{B})=qvB\sin {90}^o=qvB$ that provide a centripetal acceleration $\frac{v^2}{r}$
$\Rightarrow qvB=\frac{m{v}^2}{r}\Rightarrow mv=qBr$
$\Rightarrow$ The de-Broglie wavelength $\lambda =\frac{h}{mv}=\frac{h}{qBr}$
$\Rightarrow \frac{{\lambda }_{\alpha -particle}}{{\lambda }_{proton}}=\frac{q_{\rho }{r}_p}{q_{\alpha }{r}_{\alpha }}$
Since $\frac{r_{\alpha }}{r_p}=1$ and $\frac{q_{\alpha }}{q_p}=2$
$\Rightarrow \frac{{\lambda }_{\alpha }}{\lambda }=1/2$
Hence ($C$) is correct.