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Question
Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.
Solution
Charge on the electron, q1 = `1.6 xx 10^-19 C`
Charge on the nucleus, q2 = `1.6 xx 10^-19 C`
Let r be the distance between the nucleus and the electron.
Coulomb force (F) is given by
`F = (q_1q_2)/(4 pi∈_0r^2) ........(1)`
Here , q1 = q2 = q = 1.6`xx 10^-19C`
000 Smallest distance between the nucleus and the first orbit, r = 0.53 `r = 0.53xx10^-10m `
`K= 1/(4piepsilon_0) = 9 xx 10^9Nm^2C^-2`
Substituting the respective values in (1), we get
`F =((9xx10^9)xx(1.6xx10^-19)xx(1.6xx10^-19))/(0.53xx10^-10)^2`
= `(1.6xx1.6xx9xx10^-9)/(0.53) = 82.02 xx 10^-9`
= `8.2xx10^-8 N`
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