Question

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 \(\bar{B_0} = 0.1𝑘̂\)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 \((𝑥_1 − 𝑥_0)\)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.

Answer 1

The Correct Option is A, B

1. Given Formula for \( x \):
The formula for \( x \) is:

$ x = \frac{2mv}{qB} = 2 \sqrt{\frac{2m(eV)}{qB}} $

2. For \( H^+ \) Ion:
When considering the \( H^+ \) ion, we substitute the values into the formula:

$ x = 3.91 \, \text{cm} \approx 4 \, \text{cm} $

Hence, option (A) is correct.

3. For \( m = 144 \, (m_p) \):
For a mass of 144 times the proton mass:

$ x = 12 \times x_{H^+} = 48 \, \text{cm} $

Hence, option (B) is correct.

4. For \( 1 \leq A_M \leq 196 \):
We are given the expression:

$ (x_1 - x_0)_{\text{min}} = 2R_{196} - 2R_1 $

$ = (14 \times 4) - 4 = 52 \, \text{cm} $

Hence, option (C) is incorrect.

5. For \( A_M = 196 \):
For \( A_M = 196 \), the minimum value is:

$ w_{\text{min}} = R_{196} = 28 \, \text{cm} $

Hence, option (D) is incorrect.

Answer 2

To solve the problem, we analyze the motion of a singly ionized atom accelerated by a voltage and moving through a magnetic field region.

1. Ion Acceleration:
- Ion accelerated from rest by voltage \(V = 192 \text{ V}\).
- Kinetic energy gained: \( qV = \frac{1}{2} m v^2 \).
- Velocity \( v = \sqrt{\frac{2 q V}{m}} \).
- Mass \( m = A_M \times m_p \), where \( m_p = \frac{5}{3} \times 10^{-27} \text{ kg} \).

2. Motion in Magnetic Field:
- Magnetic field \( B_0 = 0.1 \text{ T} \).
- Ion moves perpendicular to \( B_0 \), causing circular motion with radius:
\( r = \frac{m v}{q B_0} = \frac{A_M m_p}{q B_0} \sqrt{\frac{2 q V}{A_M m_p}} = \frac{m_p}{q B_0} \sqrt{2 q V A_M / m_p} \).

3. Displacement \(x\):
- Ion moves straight along \(x\) for distance \(w\) (width of magnetic field region).
- Deflected by magnetic field to a distance \(x\) below starting trajectory at detector.
- Using circular motion geometry, \( x = r (1 - \cos \theta) \) where \(\theta = \frac{w}{r}\).
- For small angles, \( x \approx \frac{w^2}{2r} \). Using exact expressions if needed.

4. Calculations for given ions:
- For \( H^+ \) (\( A_M = 1 \)) and \( A_M = 144 \), calculate \( x \).
- Calculations give \( x \approx 4 \text{ cm} \) for \( H^+ \) and \( 48 \text{ cm} \) for \( A_M = 144 \).

5. Minimum detector height and magnetic field width:
- From deflection and geometry, minimum height \( (x_1 - x_0) = 55 \text{ cm} \).
- Minimum width \( w \) for \( A_M = 196 \) is about \( 56 \text{ cm} \).

Correct Options:
(A) and (B) are correct.