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 mh = 1.67 × 10-27 kg]

Question

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 mh = 1.67 × 10-27 kg] (A) 10-23 C  (B) 10-37 C  (C) 10-47 C  (D) 10-20 C

Tardigrade Admin · Accepted Answer

Given: m h =1.67 × 10-27 kg Net charge on hydrogen atom q =( e +Δ e )- e =Δ e Let the distance between them be d. Equating gravitational force and the electrostatic force, we get ( Gm h 2/ d 2)=( k (Δ e )2/ d 2) Or ((6.67 × 10-11)(1.67 × 10-27)2/ d 2)=((9 × 109)(Δ e )2/ d 2) Longrightarrow Δ e =2.06 × 10-37 C

$1 0^{- 23} C$

$1 0^{- 37} C$

$1 0^{- 47} C$

$1 0^{- 20} C$

10−23C

10−37C

10−47C

10−20C

Given : mh​=1.67×10−27kg Net charge on hydrogen atom q=(e+Δe)−e=Δe Let the distance between them be d. Equating gravitational force and the electrostatic force, we get d2Gmh2​​=d2k(Δe)2​ Or d2(6.67×10−11)(1.67×10−27)2​=d2(9×109)(Δe)2​ ⟹Δe=2.06×10−37C

$1 0^{- 23} C$

$1 0^{- 37} C$

$1 0^{- 47} C$

$1 0^{- 20} C$