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

Answer

C

Explanation

Net charge on one H-atom `q=-e+e+triangle=trianglee`
Net electrosatic repulsive force between two H-atoms
`F_(r)=(Kq^(2))/(d^(2))=K(trianglee)^(2)/(d^(2))`
Similarly, net gravitational attractive force between two H- atoms
" " `F_(G)=(Gm^(2))/(d^(2))`
It is given that `F_(r)-F_(G)=0`
`rArr (K(trianglee)^(2))/(d^(2))-(Gm^(2)r/(d^(2))=rArr(trianglee)^(2)=(Gm^(2))/(K)`
`(trianglee)^(2)=(6.67xx10^(-11))(1.67xx10^(-27)^(2))/(9xx10^(9))`
`rArr trianglee=1.437xx10^(-37)C`