Question

A proton, a deutron and an $\alpha -$particle are accelerated through the same potential difference and then they enter a uniform normal magnetic field. If the radius of circular path of proton is $8cm$ then the radius of circular path of deutron will be

• A. $11.31cm$
• B. $22cm$
• C. $5cm$
• D. $2.5cm$

Answer 1

Answer: (A)

$11.31cm$

A particle is accelerated through potential $V$ has a velocity $v$,

So, $\frac{1}{2}m{v}^2=qV⟹mv=\sqrt{2qmV}$

Radius of circular path, $r=\frac{\sqrt{2qmV}}{Bq}$

Thus $r\propto \sqrt{m}$ as $V$ is same

$\frac{r_d}{r_p}=\sqrt{\frac{2m}{m}}$ (given: $r_p=8cm$)

$⟹r_d=11.31cm$