Question

A proton is projected with velocity $7.45\times {10}^{5}m/s$ towards an another proton which is at rest from very large distance. The minimum distance of approach is-

• A. $5\times {10}^{-13}m$
• B. $6\times {10}^{-4}m$
• C. ${10}^{-10}m$
• D. ${10}^{-8}m$

Answer 1

Answer: (A)

$5\times {10}^{-13}m$

Consider $A$ proton of mass with speed $7.5\times {10}^{5}m/s$ directly towards another proton which is initially at rest.

we already know that $proton-proton$ repulsive force acts, both are repelling.

And consider $r$ is the distance of closest approach

Then,

$kineticenergyofproton=electrostaticenergybetweenproton$

$\begin{array}{c}\frac{1}{2}m{v}^2=\frac{K{e}^2}{r}\\ \Rightarrow r=\frac{2k{e}^2}{m{v}^2}=\frac{2\times 9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{1.675\times {10}^{-27}\times \left(7.45\times {10}^5\right)}we\end{array}$

Here, mass of proton $=1.675\times {10}^{-27}kg,e=1.6\times {10}^{-19}$

Therefore, $r=5\times {10}^{-13}m$

The distance of closet approach is $5\times {10}^{-13}m$

Hence, The Option $A$ is the correct answer.