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(B) The electrical force of attraction between the electron and proton is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.Here,$k = 9 	imes 10^

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$3.7 	imes 10^{-7} \ N$

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(B) The electrical force of attraction between the electron and proton is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.Here,$k = 9 	imes 10^9 \ N \cdot m^2/C^2$,$q_1 = q_2 = 1.6 	imes 10^{-19} \ C$,and $r = 2.5 	imes 10^{-11} \ m$.Substituting the values:$F = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19}) 	imes (1.6 	imes 10^{-19})}{(2.5 	imes 10^{-11})^2}$$F = \frac{9 	imes 2.56 	imes 10^{9 - 19 - 19}}{6.25 	imes 10^{-22}}$$F = \frac{23.04 	imes 10^{-29}}{6.25 	imes 10^{-22}}$$F = 3.6864 	imes 10^{-7} \ N \approx 3.7 	imes 10^{-7} \ N$.

In a hydrogen atom,the distance between the electron and proton is $2.5 \times 10^{-11} \ m$. The electrical force of attraction between them will be

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