QJEEAdv24Ph75c2: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph75c2 Solution: A positive, singly ionized atom of mass number is accelerated…

$\begin{aligned}& x=2 R \\& \Rightarrow x=2 \frac{P}{q B} \Rightarrow x=\frac{2 \sqrt{2 m q V}}{q B} \Rightarrow x=\frac{2}{B} \sqrt{\frac{2 m V}{q}}\end{aligned}$

Option A

For $\mathrm{H}^{+} \rightarrow \mathrm{m}=\frac{5}{3} \times 10^{-27} \mathrm{~kg}$

$\therefore x=\frac{2}{0.1} \sqrt{\frac{2 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=4 \mathrm{~cm}$

Option B

For $\mathrm{A}_{\mathrm{m}}=144$

$x=\frac{2}{0.1} \sqrt{\frac{2 \times 144 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=48 \mathrm{~cm}$

Option C

for $\mathrm{A}_{\mathrm{m}}=1$

$\mathrm{x}=4 \mathrm{~cm}$ & for $\mathrm{A}_{\mathrm{m}}=196$

$\mathrm{x}=56 \mathrm{~cm}$ .

so $\mathrm{x}_0=4 \mathrm{~cm}$ & $\mathrm{x}_1=56 \mathrm{~cm}$

$\therefore \mathrm{x}_1-\mathrm{x}_0=52 \mathrm{~cm}$ .

Option D

Minimum width $=\mathrm{R}$

for $\mathrm{A}_M=196$

$\mathrm{R}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}}$

$\mathrm{R}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}}$

$\mathrm{w}_{\min }=\mathrm{R}=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 \mathrm{~cm}$