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WBJEEA positive, singly ionized atom of mass number \(A_{\mathrm{M}}\) is accelerated from rest...
A positive, singly ionized atom of mass number \(A_{\mathrm{M}}\) is accelerated from rest by the voltage \(192 \mathrm{~V}\). Thereafter, it enters a rectangular region of width \(w\) with magnetic field \(\vec{B}_0=0.1 \hat{k}\) Tesla, as shown in the figure. The ion finally hits a detector at the distance \(x\) below its starting trajectory.
[Given: Mass of neutron/proton \(=(5 / 3) \times 10^{-27} \mathrm{~kg}\), charge of the electron \(=1.6 \times 10^{-19} \mathrm{C}\).]
Which of the following option(s) is(are) correct?
[Given: Mass of neutron/proton \(=(5 / 3) \times 10^{-27} \mathrm{~kg}\), charge of the electron \(=1.6 \times 10^{-19} \mathrm{C}\).]
Which of the following option(s) is(are) correct?
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 \(\mathrm{A}_{\mathrm{M}}=196\)
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \\
& \mathrm{R}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}} \\
& \mathrm{w}_{\min }=\mathrm{R}=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 \mathrm{~cm}
\end{aligned}$
\(\)
\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 \(\mathrm{A}_{\mathrm{M}}=196\)
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \\
& \mathrm{R}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}} \\
& \mathrm{w}_{\min }=\mathrm{R}=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 \mathrm{~cm}
\end{aligned}$
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