IIT JEE , AIEEE Forum - Physics , Electricity

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Since the electron and proton are oppositely charged

proton can be assumed to be at rest as mass of proton is 2000 time to that of electron

Now due to coulombic force of attraction electronis accelerated towards the proton and the coulombic force is varying continuously due to change in distance between electron and proton

coulombic force is given by

$F = k_C \frac{q_1 q_2}{r^2}$

where

$k_C = \frac{1}{4 \pi \epsilon_0} \approx$

= 9×10⁹ N m² C^-2 (also m F^-1) is the electrostatic constant

so work done by this force when electron attains velocity twice to that of its initial velocity then

work done by force F is given by

(1/2)mv²₂ - (1/2)mv²₁ =

^r F.dr

where m = mass of electron

here v₂ = 2v₁ and v₁ = 3.24 x 10⁵ m/sec

or (3/2)mv²₁ = ^r F.dr = kq² [-1/r]^r_infinite

or (3/2)mv²₁ = kq²/r -----(1)

here, q = 1.6 x 10^-19 C

Now substitute the values in equation (1) and find the value of r

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