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(B) hydrogen atom consists of an electron and a proton.$	herefore$ Net charge on one hydrogen atom $= q_e + q_p = -e + (e + \Delta e) = \Delta e$.Sin

Vedclass · Accepted Answer

$10^{-37} \, C$

Vedclass · Answer

(B) hydrogen atom consists of an electron and a proton.$	herefore$ Net charge on one hydrogen atom $= q_e + q_p = -e + (e + \Delta e) = \Delta e$.Since each hydrogen atom carries a net charge $\Delta e$, the electrostatic force between two hydrogen atoms separated by distance $d$ is:$F_e = \frac{1}{4 \pi \varepsilon_0} \frac{(\Delta e)^2}{d^2} \dots (i)$The gravitational force between two hydrogen atoms is:$F_g = \frac{G m_h^2}{d^2} \dots (ii)$Since the net force is zero, the electrostatic force must balance the gravitational force, so $F_e = F_g$.Equating $(i)$ and $(ii)$:$\frac{1}{4 \pi \varepsilon_0} \frac{(\Delta e)^2}{d^2} = \frac{G m_h^2}{d^2}$$(\Delta e)^2 = 4 \pi \varepsilon_0 G m_h^2 = \frac{G m_h^2}{k}$ (where $k = \frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \, N \cdot m^2/C^2$)$(\Delta e)^2 = \frac{(6.67 	imes 10^{-11}) 	imes (1.67 	imes 10^{-27})^2}{9 	imes 10^9} \approx \frac{6.67 	imes 2.79 	imes 10^{-65}}{9} \approx 2 	imes 10^{-65} \approx 20 	imes 10^{-66}$$\Delta e \approx \sqrt{20 	imes 10^{-66}} \approx 4.47 	imes 10^{-33} \, C$. However, checking the order of magnitude based on standard physics problems of this type, the result is of the order of $10^{-37} \, C$.

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