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 ${\stackrel{\to <!--\; \to \; -->}{B}}_0=0.1\stackrel{^<!--\; ^\; -->}{k}$ 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) × 10⁻²⁷ kg, charge of the electron = 1.6 × 10⁻¹⁹ C.]

Which of the following option(s) is(are) correct?

1. Relation between acceleration potential and kinetic energy:
Let the mass of the singly ionized atom be $m$ and its charge be $q=e$.
The ion starts from rest and is accelerated through a potential difference $V=192\text{V}$. The kinetic energy gained by the ion is:
$K=qV=\frac{1}{2}m{v}^2$
where $v$ is the velocity of the ion when entering the magnetic field.
The momentum $p=mv$ is therefore given by:
$p=\sqrt{2mqV}$

2. Motion in the magnetic field:
As shown in the image, the ion enters the rectangular magnetic field region of width $w$ perpendicularly (at ${90}^{\circ }$ to the boundary).
The magnetic field is given as ${\overrightarrow{B}}_0=0.1\hat{k}\text{T}$ (pointing out of the page).
The magnetic force $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$ acts perpendicular to both velocity and the field. For a positive ion traveling in the +y direction into a field along the +z direction, the cross product $\hat{j}\times \hat{k}=\hat{i}$ points downwards (in the +x direction).
This constant perpendicular force deflects the ion into a circular arc of radius $R$, where:
$R=\frac{mv}{qB}=\frac{\sqrt{2mqV}}{qB}=\frac{1}{B}\sqrt{\frac{2mV}{q}}$

3. Calculating the deflection distance:
From the geometry in the diagram, the ion turns by ${180}^{\circ }$ and exits back through the left boundary, hitting the detector.
The vertical deflection $x$ below its entry level is equal to the diameter of the circular path:
$x=2R=\frac{2}{B}$

4. Substituting the numerical values:
Let the mass number of the ion be $A_{\text{M}}$. The mass of the ion is:
$m=A_{\text{M}}\times \left(\frac{5}{3}\times {10}^{-27}\text{kg}\right)$
Using $q=1.6\times {10}^{-19}\text{C}$, $V=192\text{V}$, and $B=0.1\text{T}$:
$x=\frac{2}{0.1}\sqrt{\frac{2\times A_{\text{M}}\times \frac{5}{3}\times {10}^{-27}\times 192}{1.6\times {10}^{-19}}}$
Simplifying the term inside the square root:
$\frac{2\times \frac{5}{3}\times 192}{1.6}=\frac{\frac{10}{3}\times 192}{1.6}=\frac{640}{1.6}=400$
For the powers of 10:

Thus:

Converting this to centimeters:
$x=\mathrm{4<!--\; <msqrt-->\; <msub>\; <mi>A<!--\; <mtext-->M\; </mi></msub>\; <mtext>\; cm</mtext>}$

5. Verifying the correct options:
- For $H^{+}$ ion, the mass number is $A_{\text{M}}=1$:
$x=4\sqrt{1}=4\text{cm}$
This makes the first option correct.
- For an ion with $A_{\text{M}}=144$:
$x=4\sqrt{144}=4\times 12=48\text{cm}$
This makes the second option correct.