An electron of mass 9.1 x 10\(^{-31}\) kg moves with a speed of 2.0 x 10\(^6\) ms\(^{-1}\) round the nucleus of an atom in a Circular path of radius 6.1 x 10\(^{11}\) m. Calculate the centripetal force acting on the electron.
A. 7.7 x 10\(^{47}\) N B. 6.0 x 10\(^{-8}\)N C. 3.0 x 10\(^{-14}\)N D. 1.3 x 10\(^{-26}\)N
Correct Answer: B
Explanation
Centripetal force (Fr) = \(\frac{MV^2}{r}\) i.e. Fr = \(\frac{Mv^2}{r}\) = \(\frac{9.1 \times 10^{-31} \times 2.0 \times 10^6 \times 2.0 \times 10^6}{6.1 \times 10^{-11}}\) N = (\(\frac{36.4 \times 10^{-19}}{6.1 \times 10^{-11}}\)) N = 5.967 x 10\(^{-8}\) N ~ 6.0 x 10\(^{-8}\) N
