Question

An $\alpha -particle$ and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The $\alpha -particle$ and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths :

• A. $2:3$
• B. $3:4$
• C. $5:7$
• D. $1:2$

Answer 1

The correct option is D $1:2$

Magnetic force, $\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})=qvB\sin \theta$

As magnetic field is perpendicular to velocity so $\theta ={90}^o$ so $F_B=qvB$

or, $\frac{m{v}^2}{r}=qvB$ or $mv=qBr$

So, momentum $p=mv=qBr$

de Broglie wavelength for proton , ${\lambda }_p=\frac{h}{p_p}=\frac{h}{q_{p}B{r}_p}$ and

de Broglie wavelength for alpha particle , ${\lambda }_{\alpha }=\frac{h}{p_{\alpha }}=\frac{h}{q_{\alpha }B{r}_{\alpha }}$

Given, $\frac{r_{\alpha }}{r_p}=1$ and $q_{\alpha }=2q_p$

Thus, $\frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{r_p{q}_p}{r_{\alpha }{q}_{\alpha }}=1/2$