Question

A proton moves a speed of $7.45\times {10}^{5}m/s$ directly towards a free proton initially at rest. Find the distance of the closet approach for the two protons.

Answer 1

Speed=$7.45\times 1{-}^{5}m/s$ let mass=$m$

Let $x$ is closest distance.

$K.E$ loss=gain in potential energy

$\frac{1}{2}m{v}^2=\frac{K(9_p{)}^2}{r},q_p=e,=1.6\times {10}^{-19}=m_p=1.675\times {10}^{-27}$

=$[\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2\times 2}{1.675\times {10}^{-27}}{]}^{\frac{1}{2}}$

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