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A positive, singly ionized atom of mass number $A_M$ is accelerated from rest by the voltage $192 V$. Thereafter, it enters a rectangular region of width $w$ with magnetic field $B_0=0.1 \hat{k}$ 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 $=(5 / 3) \times 10^{-27} kg$, charge of the electron $=1.6 \times 10^{-19} C$.]
Which of the following option($s$) is(are) correct?
$(A)$ The value of $x$ for $H^{+}$ion is $4 cm$.
$(B)$ The value of $x$ for an ion with $A_M=144$ is $48 cm$.
$(C)$ For detecting ions with $1 \leq A_M \leq 196$, the minimum height $\left(x_1-x_0\right)$ of the detector is $55 cm$.
$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M=196$ is $56 cm$.
$A,B$
$A,C$
$A,D$
$A,B,C$
Solution
$x=2 R$
$\Rightarrow x =2 \frac{ P }{ qB } \Rightarrow x =\frac{2 \sqrt{2 mqV }}{ qB } \Rightarrow x =\frac{2}{ B } \sqrt{\frac{2 mV }{ q }}$
Option $A$
$\text { For } H ^{+} \rightarrow m =\frac{5}{3} \times 10^{-27} kg$
$\therefore x =\frac{2}{0.1} \sqrt{\frac{2 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=4 cm$
Option $B$
For $A_m=144$
$x=\frac{2}{0.1} \sqrt{\frac{2 \times 144 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=48 cm$
Option $C$
for $A_m=1$
$x =4 cm \&$ for $A _m=196$
$x =56 cm$.
so $x _0=4 cm \& x _1=56 cm$
$\therefore x _1- x _0=52 cm$.
Option $d$
Minimum width $=R$
for $A _{ M }=196$
$R =\frac{ P }{ qB }=\frac{\sqrt{2 mqV }}{ qB }$
$R =\frac{1}{ B } \sqrt{\frac{2 mV }{ q }}$
$W _{\min }=R=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 cm$
