Question

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. 
Assertion A : If oxygen ion (O\(^{-2}\)) and Hydrogen ion (H\(^{+}\)) enter normal to the magnetic field with equal momentum, then the path of O\(^{-2}\) ion has a smaller curvature than that of H\(^{+}\). 
Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. 
In the light of the above statements, choose the correct answer from the options given below

• A. A is true but R is false
• B. Both A and R are true but R is NOT the correct explanation of A
• C. A is false but R is true
• D. Both A and R are true and R is the correct explanation of A

Hint:

The radius of the circular path of a charged particle in a magnetic field is directly proportional to its momentum and inversely proportional to the magnitude of its charge. When comparing particles with equal momentum, the particle with a larger charge will have a smaller radius of curvature.

Answer 1

The Correct Option is A

To solve this question, we need to analyze each statement and apply relevant physics concepts.

Assertion A: If oxygen ion \((\text{O}^{2-})\) and hydrogen ion \((\text{H}^{+})\) enter normal to the magnetic field with equal momentum, then the path of \(\text{O}^{2-}\) ion has a smaller curvature than that of \(\text{H}^{+}\).

To understand this, consider the formula for the radius of curvature \((r)\) in a magnetic field:

\(r = \frac{mv}{qB}\)

Where:

• \(m\) is the mass of the ion,
• \(v\) is the velocity,
• \(q\) is the charge,
• \(B\) is the magnetic field strength.

 

Since both ions enter with equal momentum (momentum \(p = mv\)), we focus on their charges and masses. The oxygen ion \((\text{O}^{2-})\) will generally have a much larger mass compared to the hydrogen ion \((\text{H}^{+})\). Given equal momentum, the velocity of each will adjust accordingly to their mass differences. Given these differences and understanding the charges, the larger mass of the oxygen ion will result in a smaller curvature since:

\(r \propto \frac{m}{q}\text{ (if p is constant)}\)

This confirms Assertion A to be true.

Reason R: A proton with the same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.

This statement involves analyzing the path radii of a proton and an electron, which have very different masses. Given equal momentum:

\(r = \frac{p}{qB} = \frac{mv}{qB}\)

Here, a proton has a much larger mass than an electron. For equal momentum, using \(p = mv\), the electron's velocity will be higher due to its lower mass. In a magnetic field, despite having the same momentum, the larger mass of the proton causes it to have a larger radius compared to the electron due to the reciprocal relationship with mass. Thus, Reason R is false.

Therefore, the correct answer is that Assertion A is true but Reason R is false.

Answer 2

The radius of the circular path of a charged particle moving perpendicular to a uniform magnetic field is given by: \[ r = \frac{mv}{qB} = \frac{p}{qB} \] where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is the magnitude of the charge, \( B \) is the magnetic field strength, and \( p = mv \) is the linear momentum. 

For Assertion A, the oxygen ion is O\(^{-2}\), so its charge magnitude is \( |q_{O^{-2}}| = 2e \). 
The hydrogen ion is H\(^{+}\), so its charge magnitude is \( |q_{H^{+}}| = e \), where \( e \) is the elementary charge. Both ions enter the magnetic field with equal momentum \( p \). 
The radius of curvature for O\(^{-2}\) is: \[ r_{O^{-2}} = \frac{p}{2eB} \] The radius of curvature for H\(^{+}\) is: \[ r_{H^{+}} = \frac{p}{eB} \] Comparing the radii, we see that \( r_{O^{-2}} = \frac{1}{2} r_{H^{+}} \). 
This means the path of O\(^{-2}\) ion has a smaller radius of curvature (larger curvature) than that of H\(^{+}\). 
Therefore, Assertion A is true. 

For Reason R, a proton has charge \( +e \) and an electron has charge \( -e \), so their charge magnitudes are equal \( |q_p| = |q_e| = e \). They have the same linear momentum \( p \). 
The radius of curvature for the proton is: \[ r_p = \frac{p}{eB} \] The radius of curvature for the electron is: \[ r_e = \frac{p}{eB} \] Thus, \( r_p = r_e \). A proton and an electron with the same linear momentum will form paths of the same radius of curvature, not a smaller radius for the proton.

Therefore, Reason R is false. Since Assertion A is true and Reason R is false, the correct answer is (1).