Suppose the charge of a proton and an electron differ slightly. One of them is , the other is . If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance (much greater than atomic size) apart is zero, then

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Solution

Q.
Suppose the charge of a proton and an electron differ slightly. One of them is $- e$ , the other is $(e + Δ e)$ . 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 $Δ e$ is of the order of

[Given: mass of hydrogen $m_{h} = 1.67 \times 10^{- 27}$ $kg$ ] [2017]

1 $10^{- 23}$ C

2 $10^{- 37}$ C

3 $10^{- 47}$ C

4 $10^{- 20}$ C

Ans.
(2)

A hydrogen atom consists of an electron and a proton.

$∴$ $Charge on one hydrogen atom = q_{e} + q_{p} = - e + (e + Δ e) = Δ e$

Since a hydrogen atom carries a net charge $Δ e$

$∴$ $Electrostatic force, F_{e} = \frac{1}{4 π ε_{0}} \frac{{(Δ e)}^{2}}{d^{2}}$ ...(i)

will act between two hydrogen atoms.

The gravitational force between two hydrogen atoms is given as $F_{g} = \frac{G m_{h} m_{h}}{d^{2}}$ ...(ii)

Since the net force on the system is zero, $F_{e} = F_{g}$
Using eqns. (i) and (ii), we get

$\frac{{(Δ e)}^{2}}{4 π ε_{0} d^{2}} = \frac{G m_{h}^{2}}{d^{2}} or, {(Δ e)}^{2} = 4 π ε_{0} G m_{h}^{2}$

$= 6.67 \times 10^{- 11} \times {(1.67 \times 10^{- 27})}^{2} [\frac{1}{(9 \times 10^{9})}]$

$Δ e \approx 10^{- 37}$ C

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