Question

Four charge particles $\alpha$, proton, deuteron and $L{i}^{+++}$ are projected in a uniform magnetic field perpendicular to their velocity. Consider the radius of the particle having least radius out of these to be r, List-1 gives the particles and List-2 gives value of n, where nr is the radius of the particle.
Match the entries correctly in the lists if kinetic energy of each particle is same.
$\begin{array}{cccc}& \text{List-1}& & \text{List-2}\\ \text{(I)}& \alpha & \text{(P)}& 1\\ \text{(II)}& \text{Proton}& \text{(Q)}& 2\\ \text{(III)}& \text{Deuteron}& \text{(R)}& 3\\ \text{(IV)}& L{i}^{+++}& \text{(S)}& \frac{3}{\sqrt{7}}\\ & & \text{(T)}& \frac{3\sqrt{2}}{\sqrt{7}}\\ & & \text{(U)}& \frac{3}{2}\end{array}$

• A. $I\to S,II\to S,III\to T,IV\to P$
• B. $I\to P,II\to P,III\to T,IV\to Q$
• C. $I\to P,II\to P,III\to Q,IV\to S$
• D. $I\to S,II\to S,III\to P,IV\to P$

Answer 1

The correct option is A $I\to S,II\to S,III\to T,IV\to P$

$R=\frac{mv}{qB}=\frac{\sqrt{2mE}}{qB}$

$R_{\alpha }=\frac{\sqrt{2\times 4mE}}{2eB}=\frac{\sqrt{2mE}}{eB}=R_0$

$R_p=\frac{\sqrt{2mE}}{eB}=R_0$

$R_d=\frac{\sqrt{2\times 2mE}}{eB}=\frac{\sqrt{2}\sqrt{2mE}}{eB}=\sqrt{2}{R}_0$

$R_{Li}=\frac{\sqrt{2\times 7mE}}{3eB}=\frac{\sqrt{7}}{3}\frac{\sqrt{2mE}}{eB}=\frac{\sqrt{7}}{3}{R}_0$

Let, Least radius $=\frac{\sqrt{7}}{3}{R}_0=r\Rightarrow R_0=\frac{3r}{\sqrt{7}}$
Hence, $R_{\alpha }=R_0=\frac{3r}{\sqrt{7}}$
$R_p=R_0=\frac{3r}{\sqrt{7}}$
$R_d=\sqrt{2}{R}_0=\frac{3\sqrt{2}}{\sqrt{7}}r$
$R_{Li}=\frac{\sqrt{7}}{3}{R}_0=r$