Question

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

Answer 1

Correct Answer - b
Net charge on one H-atom `= -e+(e+Deltae)= DeltaE `
Net electrostatic force between two H-atoms
`=(k(Deltae)^(2))/(r^(2))` Respulsive
where `k` is `(1)/(4piepsilon_(0))= 9xx10^(9)Nm^(2)C^(-2)`
Net gravitational force between two H-atom `=(Gm^(2)H)/(r^(2))` (attractive)
Since net force on `H- atoms` is zero, therefore,
`(kxx(Deltae)^(2))/(r^(2))=(Gm^(2)H)/(r^(2))`
`Deltae=m_(H)sqrt((G)/(k))= 1.67xx10^(-27)sqrt((6.67xx10^(-11))/(9xx10^(9)))C`
`= 1.436xx10^(-37)C`
`Deltae^(2)` is one of the order of `10^(-74)` and `Deltae` is the order of `10^(-37)`