Question

Four charge particles $H{e}^{++}$, proton, deutron and $L{i}^{++}$ enters a region of uniform magnetic field and moves in a circular path of different radius. List I gives above four particles while list II gives magnitude of some quantity.

$\begin{array}{cc}\text{List I}& \text{List II}\\ \text{I -}H{e}^{++}& \text{P)}1\\ \text{II -}\text{Proton}& \text{Q)}2\\ \text{III -}\text{deutron}& \text{R)}\frac{1}{\sqrt{2}}\\ \text{IV)}L{i}^{++}& \text{S)}\sqrt{2}\\ & \text{T)}\frac{\sqrt{7}}{2}\\ & \text{U)}\frac{1}{2}\end{array}$
If kinetic energy of all four particles is same than correct match for radius of four particles in units of radius of proton is

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

Answer 1

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

$I\to P,II\to P,III\to S,IV\to T$
Since $R=\frac{mv}{qB}=\frac{\sqrt{2mE}}{qB}$ = is energy
$R_P=\frac{\sqrt{2ME}}{eB}=R_0\left(let\right)$
$R_{H{e}^{++}}=\frac{\sqrt{2\times 4mE}}{2eB}=R_0$
$R_d=\frac{\sqrt{2\times 2mE}}{eB}=\sqrt{2}{R}_0$
$R_{L{i}^{++}}=\frac{\sqrt{2\times 7mE}}{2eB}=\frac{\sqrt{7}}{2}{R}_0$