Question

.-1
17. Two electrons, each with a velocity of 106 ms
are released towards eachother What will be the
closest distance of approach between them ?​

Answer 1

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Total kinetic energy of electrons

K.E=2×12 mv2

Potential energy of electrons

U=ke2r

When they approached closest distance (r) their total kinetic energy is converted into potential energy.

K.E = U2×12 mv2=ke2r

r = ke2mv2

r = 9×109 Nm2C2×(1.6×10−19 C)29.1×10−31 kg×(106 ms)2=2.53×10−10 m

= 2.53 Å

Answer 2

Explanation:

Kinetic energy=potiental energy

2*1mv^2/2=k*e*E/v

v=k*e*E/mv^2

v=9.6*10^9*1.6*1.6*10^-19*10^-19/9.1*10^-31*10^12

v=2.53*10^-10