The zirconium nucleus contains 40 protons, and an electron is $1.0 \mathrm {~nm}$ from the nucleus. What is the electric force on the electron due to the nucleus? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \right.$. $\mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )

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The electric force $ F $ can be calculated using Coulomb's law: $ F = k \frac{|q_1 q_2|}{r^2} $. Here, $ q_1 = 40e $ (charge of the nucleus), $ q_2 = -e $ (charge of the electron), and $ r = 1.0 \times 10^{-9} $ m. Substituting the values, $ F = 9.0 \times 10^9 \times \frac{40 \times (1.6 \times 10^{-19})^2}{(1.0 \times 10^{-9})^2} \approx 9.2 $ nN.

The Zirconium Nucleus Contains 40 Protons, and an Electron Is 1.0 nm1.0 \mathrm {~nm}1.0 nm

The Zirconium Nucleus Contains 40 Protons, and an Electron Is $1.0 \mathrm {~nm}$