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

Displacement of the Particle in a Magnetic Field 

The displacement of the particle is given by:

\[ x = \frac{2R}{\frac{2mv}{qB}} \] where \( R \) is the radius of the circular path, \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is its charge, and \( B \) is the magnetic field strength.

Substituting \( v = \frac{2qV}{m} \), we get:

\[ x = \frac{2 \cdot m \cdot \frac{2qV}{m}}{qB} \] Simplifying: \[ x = \frac{4m(qV)}{qB} \] For a proton (\( H^+ \)): \[ x = 4 \, \text{cm} \]

For a Particle with Atomic Mass \( \text{AM} = 144 \):

The displacement is proportional to the square root of the mass: \[ x \propto \sqrt{m} \] Thus, for \( \text{AM} = 144 \) (144 times the proton mass): \[ x = 4 \, \text{cm} \cdot \sqrt{144} = 4 \cdot 12 = 48 \, \text{cm} \]

Conclusion:

The correct options are:

• Options (1) and (2)