Question

Two electrons are moving towards each other each with a velocity of $ {10^6}m/s $ . What will be the closest distance of approach between them?
(A) $ 1.53 \times {10^{ - 8}}m $
(B) $ 2.53 \times {10^{ - 10}}m $
(C) $ 2.53 \times {10^{ - 6}}m $
(D) $ zero $

Answer 1

Hint: With the decrease in kinetic energy, the potential energy will increase. knowing this we will put the kinetic energy of two electrons equal to their potential energy. then we will solve for the closest distance of approach between them.
The total kinetic energy of electrons, $ Kinetic\;Energy = \dfrac{1}{2}m{\nu ^2} $
The potential energy of electrons, $ Potential\;Energy = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{r} $
 $ \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\dfrac{{N{m^2}}}{{{C^2}}} $
 $ e = 1.6 \times {10^{ - 19}}C $
 $ m = 9.1 \times {10^{ - 31}}kg $ .

Complete answer:

Two electrons of the same mass are moving towards each other with the same velocity $ \nu = {10^6}m/s $ . So at the same time, they will come to zero, when the distance between them is about $ r $ which is the closest distance of approach between them then kinetic energy will be zero.
Therefore, for two electrons, When they approach the closest distance the total kinetic energy of the two electrons will get converted into potential energy of electrons. Hence, the closest distance of approach is when its total kinetic energy is converted into potential energy.
Gain in potential energy $ = $ Loss in kinetic energy
 $ \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{r} = \dfrac{1}{2}m{\nu ^2} + \dfrac{1}{2}m{\nu ^2} $
Since, $ \left( {{q_1} = {q_2} = e} \right) $
 $ e = 1.6 \times {10^{ - 19}}C $ is the charge of electron
 $ r = $ is the distance between the electrons
 $ m = 9.1 \times {10^{ - 31}}kg $ is the mass of electron
 $ \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\dfrac{{N{m^2}}}{{{C^2}}} $
 $ \nu = $ is the velocity of electrons
 $ \Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{r} = m{\nu ^2} $
For finding the closest distance of approach between them,
 $ \Rightarrow r = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{m{\nu ^2}}} $
Putting all the values from above
 $ \Rightarrow r = \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}}} $
 $ \Rightarrow r = 2.53 \times {10^{ - 10}}m $
Hence option B) $ 2.53 \times {10^{ - 10}}m $ is the correct option.

Note:
The energy held in an object as a result of its position or arrangement is known as potential energy. Kinetic energy is the energy released by an object as a result of its movement, or motion. Kinetic energy can be turned into potential energy, and potential energy can be converted back into kinetic energy.