Question

Two electrons, each moving with a velocity of ${10}^6\text{m/s}$ towards each other. The closest distance of approach between them is
$($Charge on an electron, $e=-1.6\times {10}^{-19}\text{C}$ and mass of an electron, $m_e=9.1\times {10}^{-31}\text{kg})$

• A. $2.5\times {10}^{-10}\text{m}$
• B. $4.5\times {10}^{-10}\text{m}$
• C. $3.5\times {10}^{-10}\text{m}$
• D. $5.5\times {10}^{-10}\text{m}$

Answer 1

The correct option is A $2.5\times {10}^{-10}\text{m}$

Given that,
Velocity of the electron, $v={10}^6\text{m/s}$
Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

The electrons are negatively charged. When the electrons approach each other, they will experience retarding electrostatic force and their velocity will start decreasing. Their velocity will be zero at closest approach.

Let the closest distance of approach is $r$.

On applying law of conservation of mechanical energy,
Loss in kinetic energy $=$ Gain in potential energy
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=\frac{k{e}^2}{r}$
$\Rightarrow r=\frac{k{e}^2}{m{v}^2}$
$\Rightarrow r=\frac{9\times {10}^9\times {(1.6\times {10}^{-19})}^2}{9.1\times {10}^{-31}\times ({10}^6{)}^2}$

$\Rightarrow r=2.5\times {10}^{-10}\text{m}$