Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + $Δe$ ) . If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then $Δe$ is of the order of [Given mass of hydrogen $m_{h} = 1 .67 \times 10^{- 27} kg$ ]

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$10^{- 37} C$

$10^{- 47} C$

$10^{- 20} C$

$10^{- 23} C$

answer is D.

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Detailed Solution

$Net charge on hydrogen atom q = (e + Δe) - e = Δe$

$Equating gravitational force and the electrostatic force, we get$

$\frac{{Gm}_{h}^{2}}{d^{2}} = \frac{k (Δ e)^{2}}{d^{2}}$

$\Rightarrow \frac{(6.67 \times 10^{- 11}) {(1.67 \times 10^{- 27})}^{2}}{d^{2}} = \frac{(9 \times 10^{9}) (Δ e)^{2}}{d^{2}}$

$\Rightarrow Δ e = 2.06 \times 10^{- 37} C$

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