Question

The ratio of de-Broglie wavelength of a $\alpha$ -particle to that of a proton when they are subjected to the same magnetic field and their radii in the magnetic field are equal (assume the field induction vector $\overrightarrow{B}$ is perpendicular to the velocity vectors of the $\alpha$ -particle and the proton) is

• A. $1$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

When a charged particle (charge q, mass m) enters perpendicularly in a magnetic field (B) than, radius of the path described by it $r=\frac{mv}{qB}\Rightarrow mv=qBr$.
Also de-Broglie wavelength $\lambda =\frac{h}{mv}$
$\Rightarrow \lambda =\frac{h}{qBr}\Rightarrow \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{q_p{r}_p}{q_{\alpha }{r}_{\alpha }}=\frac{1}{2}$