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(B) According to Coulomb's law,the electrostatic force of attraction $(F)$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is gi

Vedclass · Accepted Answer

$\frac{e^2}{r^2}$

Vedclass · Answer

(B) According to Coulomb's law,the electrostatic force of attraction $(F)$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula: $F = k \frac{|q_1 q_2|}{r^2}$.In a hydrogen atom,the nucleus has a charge of $+e$ (where $Z=1$) and the electron has a charge of $-e$.Substituting these values into the formula,we get: $F = k \frac{|(+e)(-e)|}{r^2} = k \frac{e^2}{r^2}$.Assuming the constant $k$ is taken as unity in the given options,the expression simplifies to $\frac{e^2}{r^2}$.

The electrostatic force of attraction between the electron and the nucleus in a hydrogen atom is given by:

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