Q. Two tree protons are separated by a distance of 1$\mathring{\mathrm{A}}$. If they are released, the kinetic energy of each proton when at infinite separation is
A) $11.5 \times 10^{- 19\ }$J
B) $23 \times 10^{- 19\ }$J
C) $46 \times 10^{- 19\ }$J
D) $5.6 \times 10^{- 12\ }$J
Explanation
(a) KE = PE of two protons
$= \frac{e^{2}}{4\pi\varepsilon_{0}r} = \frac{9 \times 10^{9} \times \left( 1.6 \times 10^{- 19} \right)^{2}}{10^{- 10}}$
$= \ 23 \times 10^{- 19}$ J
$\therefore\ \ $KE of each proton $= \frac{23}{2} \times 10^{- 19}\ J\ $
$= 11.5 \times 10^{- 19\ }J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.