Question 7

1. Kinetic Energy and Velocity of the Ion

A positive, singly ionized atom has a charge of $q = e = 1.6 \times 10^{-19} \text{ C}$. Its mass can be expressed in terms of its mass number $A_M$: $m = A_M \times m_p = A_M \times \left(\frac{5}{3} \times 10^{-27}\right) \text{ kg}$ When accelerated from rest through a potential difference $V = 192 \text{ V}$, the kinetic energy gained is: $K = qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}}$

2. Radius of the Circular Path

Upon entering the magnetic field $\vec{B}_0 = 0.1\hat{k} \text{ T}$ perpendicularly, the ion moves in a circular path. The radius $R$ of this path is given by: $R = \frac{mv}{qB_0} = \frac{1}{B_0}\sqrt{\frac{2mV}{q}}$ Substituting the given values into the equation: $R = \frac{1}{0.1} \sqrt{\frac{2 \times \left(A_M \times \frac{5}{3} \times 10^{-27}\right) \times 192}{1.6 \times 10^{-19}}}$ $R = 10 \sqrt{A_M \times \frac{640}{1.6} \times 10^{-8}} = 10 \times 10^{-4} \sqrt{400 A_M}$ $R = 10^{-3} \times 20\sqrt{A_M} \text{ m} = 0.02\sqrt{A_M} \text{ m} = 2\sqrt{A_M} \text{ cm}$

3. Finding the Deflection $x$

As shown in the figure, the ion enters normally ($90^\circ$) to the boundary of the magnetic field region, completes a semi-circle ($180^\circ$ deflection), and exits the same boundary moving horizontally to the left. Because it exits moving perfectly horizontally, the vertical distance $x$ below its starting trajectory is exactly equal to the diameter of its circular path:

$x = 2R = 4\sqrt{A_M} \text{ cm}$

Evaluating the Options

• Option (A): Value of $x$ for $\text{H}^+$ ion ($A_M = 1$) $x = 4\sqrt{1} = \mathbf{4 \text{ cm}}$

• Option (A) is correct.

• Option (B): Value of $x$ for an ion with $A_M = 144$ $x = 4\sqrt{144} = 4 \times 12 = \mathbf{48 \text{ cm}}$

• Option (B) is correct.

• Option (C): Minimum height $(x_1 - x_0)$ of the detector for $1 \le A_M \le 196$ For $A_M = 1$: $x_{\text{min}} = 4\sqrt{1} = 4 \text{ cm}$ For $A_M = 196$: $x_{\text{max}} = 4\sqrt{196} = 4 \times 14 = 56 \text{ cm}$ The minimum height required to capture all these ions is: $x_1 - x_0 = 56 - 4 = \mathbf{52 \text{ cm}}$

  • Option (C) states 55 cm, so it is incorrect.

• Option (D): Minimum width $w$ of the magnetic field region for $A_M = 196$ For the ion to complete its turn and exit without hitting the opposite boundary, the width $w$ must be at least equal to its radius $R$. For $A_M = 196$: $w_{\text{min}} = R = 2\sqrt{196} = \mathbf{28 \text{ cm}}$

  • Option (D) states 56 cm, so it is incorrect.