Question

Find the ratio of de-Broglie wavelength of an $\alpha -$particle to that of a proton. The proton is subjected to double the strength of magnetic field that of $\alpha -$particle so that the radii of their path are equal to each other. Assume the field induction $\overrightarrow{B}$ perpendicular to velocity vector of the $\alpha -$particle.

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is A $1:1$

As, radius of path

$r=\frac{mV}{qB}$

$\therefore mV=qBr$

De-Broglie wavelength is given by

$\lambda =\frac{h}{mV}$

$\therefore \lambda \propto \frac{1}{mV}$

$\Rightarrow \lambda \propto \frac{1}{qBr}$

$\therefore \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{q_p{B}_p{r}_p}{q_{\alpha }{B}_{\alpha }{r}_{\alpha }}$

Given $:B_p=2B_{\alpha }\text{and}{r}_p=r_{\alpha }$

Also, $q_p=+e,q_{\alpha }=+2e$

$\therefore \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{(+e)\left(2B\right)\times r}{(+2e)\left(B\right)\times r}=\frac{1}{1}$

Hence, option $\left(\text{A}\right)$ is correct.