Atomic StructureHard

Question

If an electron is revolving around the nucleus of He+ ion at a distance of 4.0 Å, the magnitude of centripetal force on electron by the nucleus is (e = 1.6 × 10−19 C)

Options

A.2.88 × 10−9 N
B.2.88 × 10−7 N
C.1.152 × 10−18 N
D.1.44 × 10−9 N

Solution

$F = \frac{1}{4\pi\varepsilon_{0}}.\frac{ze^{2}}{r^{2}} = 9 \times 10^{9} \times \frac{2 \times \left( 1.6 \times 10^{- 19} \right)^{2}}{\left( 4 \times 10^{- 10} \right)^{2}} = 2.88 \times 10^{- 9}\text{ N}$

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