Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Two singly ionized isotopes of an element are accelerated from rest through the same potential difference and they enter perpendicular into a uniform magnetic field.
Column-I | Column-II |
(i) Their respective KE before entering into magnetic field | [A] Remains constant |
(ii) Their respective KE during motion in magnetic field | [B] Are equal |
(iii) In the magnetic field path traced is | [C] Straight line |
(iv) In the magnetic field they cannot trace a path in | [D] Circular arc |
Correct Matrix Matching
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: When charged particles are accelerated through a potential difference, they gain kinetic energy equal to the work done on them by the electric field. For two singly ionized isotopes accelerated from rest through the same potential difference, the kinetic energy (KE) is given by the equation:
$$ KE = qV $$
where q is the charge and V is the potential difference. Since both isotopes are singly ionized, they have the same charge (e) and are subjected to the same potential difference. Therefore, they will have the same amount of kinetic energy upon entering the magnetic field.
Step 2: In the uniform magnetic field, the force acting on the charged particles is perpendicular to their velocity, resulting in circular motion. The kinetic energy remains constant since there are no non-conservative forces doing work on the system.
Therefore, the kinetic energy just before entering the magnetic field remains constant.
Thus, the answer for part (i) is that their respective KE remains constant.
$$ KE = qV $$
where q is the charge and V is the potential difference. Since both isotopes are singly ionized, they have the same charge (e) and are subjected to the same potential difference. Therefore, they will have the same amount of kinetic energy upon entering the magnetic field.
Step 2: In the uniform magnetic field, the force acting on the charged particles is perpendicular to their velocity, resulting in circular motion. The kinetic energy remains constant since there are no non-conservative forces doing work on the system.
Therefore, the kinetic energy just before entering the magnetic field remains constant.
Thus, the answer for part (i) is that their respective KE remains constant.
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