Which of the following option(s) is(are) correct?A positive, singly ionized atom of mass number $A_{\mathrm{M}}$ is accelerated from rest by the voltage $192…
Which of the following option(s) is(are) correct?- The value of $x$ for $H^{+}$ion is $4 \mathrm{~cm}$.
- The value of $x$ for an ion with $A_{\mathrm{M}}=144$ is $48 \mathrm{~cm}$.
- For detecting ions with $1 \leq A_{\mathrm{M}} \leq 196$, the minimum height $\left(x_1-x_0\right)$ of the detector is $55 \mathrm{~cm}$.
- The minimum width $w$ of the region of the magnetic field for detecting ions with $A_{\mathrm{M}}=196$ is $56 \mathrm{~cm}$.
Solution
$
\begin{aligned}
& \mathrm{x}=2 \mathrm{R} \\
& \Rightarrow \mathrm{x}=2 \frac{\mathrm{P}}{\mathrm{qB}} \Rightarrow \mathrm{x}=\frac{2 \sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \Rightarrow \mathrm{x}=\frac{2}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{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 $\mathrm{A}_m=1$
$\mathrm{x}=4 \mathrm{~cm} \&$ 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 $=R$
for $\mathrm{A}_{\mathrm{M}}=196$
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \\
& \mathrm{R}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}} \\
& \mathrm{w}_{\min }=\mathrm{R}=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 \mathrm{~cm}
\end{aligned}$Asked in: JEE Advanced 2024 (Paper 2)
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