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 resolve this problem, each statement requires analysis using applicable physics principles.

Assertion A: When an oxygen ion \((\text{O}^{2-})\) and a hydrogen ion \((\text{H}^{+})\) enter a magnetic field perpendicularly with identical momentum, the trajectory of the \( \text{O}^{2-} \) ion exhibits less curvature than that of the \( \text{H}^{+} \) ion.

The radius of curvature \( (r) \) in a magnetic field is described by the formula:

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

Where:

• \( m \) denotes the ion's mass,
• \( v \) represents its velocity,
• \( q \) signifies its charge,
• \( B \) is the magnetic field strength.

Given that both ions possess equal momentum (where momentum \( p = mv \)), the analysis focuses on their respective charges and masses. The oxygen ion \( (\text{O}^{2-}) \) typically has a significantly greater mass than the hydrogen ion \( (\text{H}^{+}) \). With equal momentum, the velocity of each ion will adjust according to their mass disparity. Considering these differences and their charges, the larger mass of the oxygen ion will result in a reduced curvature, as demonstrated by the relationship:

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

This observation validates Assertion A.

Reason R: A proton possessing the same linear momentum as an electron will follow a path with a smaller radius of curvature upon perpendicular entry into a uniform magnetic field.

This statement necessitates an examination of the path radii for a proton and an electron, which differ substantially in mass. With equal momentum:

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

The proton's mass is considerably greater than the electron's. For equal momentum, and given \( p = mv \), the electron's velocity will be higher due to its lesser mass. Within a magnetic field, despite their shared momentum, the proton's greater mass leads to a larger radius of curvature compared to the electron, owing to the inverse relationship with mass. Consequently, Reason R is incorrect.

Therefore, the accurate conclusion is that Assertion A is true and Reason R is false.