Question

An $\alpha$-particle and a proton are accelerated from rest through same potential difference and both enter into a uniform perpendicular magnetic field. Find the ratio of their radii of curvature.

• A. $\sqrt{4}:1$.
• B. $\sqrt{2}:1$.
• C. $\sqrt{3}:1$.
• D. $\sqrt{2}:2$.

Answer 1

Answer: (B)

$\sqrt{2}:1$.

$\begin{array}{c}\text{Given - * Two particles}\alpha \text{particle and a proton accelerated with same potential in}\\ \text{a uniform magnetic field.}\end{array}$

$\begin{array}{cc}\Rightarrow qv=\frac{1}{2}m{v}^2& \Rightarrow q=\text{charge}\\ & v=\text{Potential}\\ & m=\text{mass}\\ & v=\text{velocity}\text{---- eqn 1}\end{array}$

$\text{we know that in a magnetic field}$

$r=\frac{mv}{Bq}-\left(2\right)\text{where}r=\text{radius of curvature}$

$\text{From relation}-\left(1\right)\text{and}\left(2\right)\text{we get ,}$

$r=\frac{1}{B}\cdot \sqrt{\frac{2mV}{q}}$

$\begin{array}{cc}& \Rightarrow \frac{r_{\alpha }}{r_p}=\sqrt{\frac{m_{\alpha \cdot q_p}}{m_p\cdot q_{\alpha }}}\begin{array}{c}\alpha \to \text{alpha particle}\\ p\to \text{proton}\end{array}\\ & \text{we know ,}{m}_{\alpha }=4m_p=1\\ & q_{\alpha }=2q_p=1\end{array}$

$\Rightarrow \frac{r_{\alpha }}{r_p}=\frac{\sqrt{2}}{1}\text{Ratio}=\sqrt{2}:1$