Q. In a hydrogen atom, the distance between the electron and proton is $2.5 \times 10^{- 11}m$. The electrical force of attraction between them will be
A) $2.8 \times 10^{- 7}N$
B) $3.7 \times 10^{- 7}N$
C) $6.2 \times 10^{- 7}N$
D) $9.1 \times 10^{- 7}N$
Explanation
(b) $F = kq_{1}\ q_{2}/r^{2},i.e.,$
$F = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{\left( 2.5 \times 10^{- 11} \right)^{2}} = 3.7 \times 10^{- 7}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.