Question

31. In a field free region, two electrons are released to
move on a line towards each other with veloaties
10^6 ms. The distance of their closest approach will
be nearer to
(a) 1.28 < 10-14 m (b) 1.92 x 10m
(c) 2.56 * 10-10 m
(d) 3.84 * 10-19 m​

Answer 1

Answer: As the electrons are moving towards each other they will slow down due to their mutual repulsive force. The potential energy of the electron system will increase and kinetic energy will be decreasing so at the minimum point of their approach the potential energy will have the maximum value and the kinetic energy will be zero but the energy is conserved so the amount of potential energy finally achieved will be equal to the amount of initial kinetic energy that was in the system putting them equal you will get the answer.

Option C is the answer.

For solution you May see attachment.

Answer 2

Answer:

0.51nm

Explaination:

Let r be distance of closest approach.

v=106ms−1

∴ Kinetic energy = Potential energy between the two electrons

or 12mev2=14πe0=q1q2r

r=2q1q24πe0×1mev2

=2×(1.6×10−19)2×9×1099.1×10−31×(106)2

=5.1×10−10=0.51×10−9=0.51nm