$\Rightarrow x =2 \frac{ P }{ qB } \Rightarrow x =\frac{2 \sqrt{2 mqV }}{ qB } \Rightarrow x =\frac{2}{ B } \sqrt{\frac{2 mV }{ q }}$

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 $A _{ M }=196$

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

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

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