JEE Advanced 2024 — Paper 2 — PHY-II solution

JEE Advanced2024Paper 2PHY-II

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 $= (5/3) \times 1 0^{- 27}$ kg, charge of the electron $= 1.6 \times 1 0^{- 19}$ 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 ( $x_{1} - x_{0}$ ) 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.

Solution

After acceleration through potential $V$ , the ion's speed satisfies $\frac{1}{2} m v^{2} = e V$ , so $v = 2 e V / m .$

Inside the field, the ion follows a circular arc of radius $R = \frac{m v}{q B _{0}} = \frac{2 m e V}{e B _{0}} .$

If the ion enters perpendicular to one edge of a sufficiently wide field region and re-emerges from the same edge, the perpendicular displacement on exit is $x = 2 R$ .

For $H^{+}$ (mass $m = \frac{5}{3} \times 1 0^{- 27}$ kg):

$R = \frac{1}{e B _{0}} 2 m e V = \frac{1}{1.6 \times 1 0 ^{- 19} \cdot 0.1} 2 \cdot \frac{5}{3} \times 1 0^{- 27} \cdot 1.6 \times 1 0^{- 19} \cdot 192 \approx 2 cm .$

Hence $x = 2 R \approx 4$ cm. (A) is correct.

Since $R \propto m \propto A_{M}$ , for $A_{M} = 144 = 1 2^{2}$ the radius is $12$ times that of $H^{+}$ , giving $x = 12 \times 4 = 48$ cm. (B) is correct.

For $1 \leq A_{M} \leq 196$ : minimum height of detector $= 2 R_{196} - 2 R_{1} = (14 - 1) \times 4 = 52$ cm, not $55$ cm. (C) is incorrect.

For $A_{M} = 196$ , the minimum width to allow full semicircular turnaround is $w = R_{196} = 28$ cm, not $56$ cm. (D) is incorrect.

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