Question

A proton and an $\alpha$ - particle accelerated through the same potential difference enter a region of uniform magnet field normally. If the radius of proton orbit is 10 cm then radius of $\alpha$ - particle orbit is

• A. $10\sqrt{2}cm$
• B. $20\sqrt{2}cm$
• C. $5\sqrt{2}cm$
• D. $7\sqrt{2}cm$

Answer 1

The correct option is A $10\sqrt{2}cm$

$\frac{m{v}^2}{r}=Bqv;r=\frac{mv}{qB}$
When a charged partice of mass m and charge q is accelerated through a potential difference V, then:
$\Rightarrow \frac{1}{2}m{v}^2=qV$
So, v = $\sqrt{\frac{2Vq}{m}}$

$r=\frac{m}{Bq}\sqrt{\frac{2Vq}{m}};r\alpha \sqrt{\frac{m}{q}}\frac{r_{\alpha }}{r_p}=\sqrt{\frac{m_{\alpha }\times q_p}{m_p\times q_{\alpha }}}=\sqrt{\frac{4}{1}\times \frac{1}{2}}=\sqrt{2}{r}_{\alpha }=10\sqrt{2}cm$