Question

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?

• A. $1.53\cdot \times {10}^{-8}m$
• B. $2.53\cdot \times {10}^{-10}m$
• C. $2.53\cdot \times {10}^{-6}m$
• D. Zero

Answer 1

Answer: (B)

$2.53\cdot \times {10}^{-10}m$

$K.E.=P.E.$

$2\times \frac{1}{2}m{v}^2=\frac{k\times e\times e}{r}$

$r=\frac{k\times e^2}{m{v}^2}=\frac{9\times {10}^9\times {(1.6\times {10}^{-19})}^2}{9.1\times {10}^{-31}\times {\left({10}^6\right)}^2}$

$r=2.53\times {10}^{-29+19}$

$r=2.53\times {10}^{-10}m$