A positive, singly ionized atom of mass number 𝐴M is accelerated from rest by the voltage 192 V. Thereafter, it enters a rectangular region of width 𝑤 with magnetic field 𝐵⃗ 0 = 0.1𝑘̂ 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? (A) The value of 𝑥 for 𝐻 + ion is 4 cm. (B) The value of 𝑥 for an ion with 𝐴M = 144 is 48 cm. (C) For detecting ions with 1 ≤ 𝐴M ≤ 196, the minimum height (𝑥1 − 𝑥0) of the detector is 55 cm. (D) The minimum width 𝑤 of the region of the magnetic field for detecting ions with 𝐴M = 196 is 56 cm.

 To solve this problem, we need to analyze the motion of a singly ionized atom in a magnetic field after it has been accelerated by a voltage. The ion follows a circular path in the magnetic field, and the distance 

$x$ at which it hits the detector depends on its mass and charge.

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Step 1: Velocity of the Ion After Acceleration

The ion is accelerated from rest by a voltage $V=192\text{V}$. The kinetic energy gained by the ion is:

$$\frac{1}{2}m{v}^2=qV$$

where:

• $m$ is the mass of the ion,

• $v$ is its velocity,

• $q=1.6\times 10^{-19}\text{C}$ is the charge of the ion.

Solving for $v$:

$$v=\sqrt{\frac{2qV}{m}}$$

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Step 2: Radius of the Circular Path in the Magnetic Field

The ion enters a region with a magnetic field $B_0=0.1\text{T}$. The magnetic force provides the centripetal force for circular motion:

$$qv{B}_0=\frac{m{v}^2}{r}$$

Solving for the radius $r$:

$$r=\frac{mv}{q{B}_0}$$

Substitute $v=\sqrt{\frac{2qV}{m}}$:

$$r=\frac{m}{q{B}_0}\cdot \sqrt{\frac{2qV}{m}}=\sqrt{\frac{2mV}{q{B}_0^2}}$$

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Step 3: Distance $x$ at Which the Ion Hits the Detector

The ion follows a semicircular path in the magnetic field, so the distance $x$ is equal to the diameter of the circular path:

$$x=2r=2\cdot \sqrt{\frac{2mV}{q{B}_0^2}}$$

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Step 4: Mass of the Ion

The mass of the ion is given by:

$$m=A_M\cdot m_p$$

where:

• $A_M$ is the mass number of the ion,

• $m_p=\frac{5}{3}\times 10^{-27}\text{kg}$ is the mass of a proton/neutron.

Substitute $m=A_M\cdot \frac{5}{3}\times 10^{-27}\text{kg}$ into the expression for $x$:

$$x=2\cdot \sqrt{\frac{2\cdot A_M\cdot \frac{5}{3}\times 10^{-27}\cdot 192}{1.6\times 10^{-19}\cdot (0.1{)}^2}}$$

Simplify:

$$x=2\cdot \sqrt{\frac{2\cdot A_M\cdot \frac{5}{3}\cdot 192\times 10^{-27}}{1.6\times 10^{-19}\cdot 0.01}}$$$$x=2\cdot \sqrt{\frac{2\cdot A_M\cdot \frac{5}{3}\cdot 192\times 10^{-27}}{1.6\times 10^{-21}}}$$$$x=2\cdot \sqrt{\frac{2\cdot A_M\cdot \frac{5}{3}\cdot 192}{1.6}\times 10^{-6}}$$$$x=2\cdot \sqrt{A_M\cdot \frac{5}{3}\cdot 240\times 10^{-6}}$$$$x=2\cdot \sqrt{A_M\cdot 400\times 10^{-6}}$$$$x=2\cdot \sqrt{A_M}\cdot 20\times 10^{-3}$$$$x=0.04\cdot \sqrt{A_M}\text{m}$$$$x=4\cdot \sqrt{A_M}\text{cm}$$

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Step 5: Evaluate the Options

(A) The value of $x$ for $H^{+}$ ion is 4 cm.

• For $H^{+}$, $A_M=1$:

  $$x=4\cdot \sqrt{1}=4\text{cm}$$

  This matches the given value. Hence, (A) is correct.

(B) The value of $x$ for an ion with $A_M=144$ is 48 cm.

• For $A_M=144$:

  $$x=4\cdot \sqrt{144}=4\cdot 12=48\text{cm}$$

  This matches the given value. Hence, (B) is correct.

(C) For detecting ions with $1\le A_M\le 196$, the minimum height $(x_1-x_0)$ of the detector is 55 cm.

• For $A_M=1$, $x_0=4\text{cm}$.

• For $A_M=196$, $x_1=4\cdot \sqrt{196}=4\cdot 14=56\text{cm}$.

• The height difference is:

  $$x_1-x_0=56-4=52\text{cm}$$

  This does not match the given value. Hence, (C) is incorrect.

(D) The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M=196$ is 56 cm.

• The width $w$ must be at least equal to the diameter of the circular path for $A_M=196$:

  $$w=x=56\text{cm}$$

  This matches the given value. Hence, (D) is correct.

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Final Answer:

The correct options are:

$$\boxed{{\displaystyle \text{(A), (B), and (D)}}}$$