Question

Two electrons are moving toward each other each with a velocity of ${10}^6$ m/s what will be the closest distance of approach between them.

Answer 1

Step 1, Given data

The velocity of the electrons = ${10}^6$ m/s

Step 2, Finding the closest distance

From the above figure, we can say that the energy of the system will be conserved.

From the energy conservation

We can write,

$\frac{1}{2}{m}_e{v}^2+\frac{1}{2}{m}_e{v}^2=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{d}$

Or, $m_e{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{d}$

Or, Distance can be given as $d=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{{m{v}^2}_e}$

Now putting all the values

$d=\frac{9\times {10}^9\times {1.6}^2\times {\left({10}^{-19}\right)}^2}{9.11\times {10}^{-31}\times {\left({10}^6\right)}^2}$

After solving

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

Hence the closest distance is $d=2.53\times {10}^{-10}m$