Question

Four charge particles $\alpha$, proton, deuteron and $L{i}^{+++}$ are projected in a uniform magnetic field perpendicular to 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 momentum 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. $I\to U,II\to R,III\to P,IV\to Q$
• B. $I\to Q,II\to P,III\to P,IV\to R$
• C. $I\to U,II\to R,III\to R,IV\to P$
• D. $I\to P,II\to Q,III\to Q,IV\to P$

Answer 1

The correct option is C $I\to U,II\to R,III\to R,IV\to P$

$R=\frac{mv}{qB}=\frac{p}{qB}$

$R_{\alpha }=\frac{p}{2eB}=\frac{R_0}{2}$

$R_p=\frac{p}{eB}=R_0$

$R_d=\frac{p}{eB}=R_0$

$R_{Li}=\frac{p}{3eB}=\frac{R_0}{3}$

$R_{Li}=\frac{R_0}{3}$ is least, let it be r

$\Rightarrow R_p=R_d=R_0=3r$ and $R_{\alpha }=\frac{R_0}{2}=\frac{3r}{2}$