Suppose the charge of a proton and an electron differ slightely. One of them is `-e`, the other is `(e+Deltae)`. 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 `Deltae` is of the order of [Given mass of hydrogen `m_(h)=1.67xx10^(-27)kg`]

Options

A. `10^(-23) C`
B. `10^(-37)C`
C. `10^(-47)C`
D. `10^(-20)C`

Answer

B

Explanation

A hydrogen atom consists of an electron and a proton.
`:.` Charge on one hydrogen atom
`= q_(e) + q_(p) = -e + (e+ Delta e) = Delta e`
Since a hydrogen atom carry a net charge `Delta e`,
`:.` Electrostatic force, `F_(e) = (1)/(rpi epsilon_(0))((Delta e)^(2))/(d^(2))` ...(i)
will act between two hydrogen atoms.

The gravitational force between two hydrogen atoms is given as
`F_(g) = (Gm_(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
`((Delta e)^(2))/(4pi epsilon_(0)d^(2)) = (Gm_(h)^(2))/(d^(2))`
`(Delta e)^(2) = 4pi epsilon_(0) Gm_(h)^(2)`
`= 6.67 xx 10^(-11) xx (1.67 xx 10^(-27))^(2)/(9 xx 10^(9))`
`Delta e ~~ 10^(-37)C`