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

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).