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.
6.1 x 10\(^{11}\) m. Calculate the centripetal force acting on the electron.
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Correct Answer: Option 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
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