Question

Two electrons are approaching each other with a relative velocity of $10^6 \, m \, s^{-1}$ . The closest distance of their approach is

• A. $ 0.51 \, nm $
• B. $ 0.51 \, mm $
• C. $ 0.51 \mathring A $
• D. $0.51 \, \mu m$

Answer 1

Answer: (A)

Let r be distance of closest approach.
$v = 10^6 \, m \, s^{-1}$
$\therefore $ Kinetic energy = Potential energy between the two electrons
or $ \frac{1}{2} m_{e} v^{2} = \frac{1}{4 \pi e_{0}} = \frac{q_{1}q_{2}}{r} $
$r = \frac{2 q_{1}q_{2}}{4 \pi e_{0}} \times\frac{1}{m_{e} v^{2}}$
$ = \frac{2 \times \left(1.6 \times 10^{-19}\right)^{2}\times 9 \times 10^{9}}{9.1 \times 10^{-31}\times \left(10^{6}\right)^{2}} $
$ =5.1 \times 10^{-10} =0.51 \times 10^{-9} = 0.51 nm$