Explanation
Two electrons are moving towards each other each with a velocity of $${\text{1}}{{\text{0}}^{{\text{6}}\;}}{\text{m/s}}$$.What will be closest distance of approach between them?
$$K.E.=P.E.$$
$$2 \times \dfrac{1}{2} mv^{2}=\dfrac{k \times e \times e}{r}$$
$$r=\dfrac{k \times e^{2}}{m v^2}=\dfrac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^2}{9.1 \times 10^{-31} \times\left(10^{6}\right)^{2}}$$
$$r=2.53 \times 10^{-29+19}$$
$$r=2.53 \times 10^{-10} \mathrm{m}$$
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