Question : Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen, mh=1.67×10-27 kg] [NEET 2017]
a) 10-20C
b) 10-23C
c) 10-37C
d) 10-47C
Doubt by Amaan (2021) and Tannu (2023)
Solution :
As per the questions,
Let the charge on each electron would be (-e) while the charge on each proton would be (e+Δe).
We know, each hydrogen atom has 1 electron, 1 proton and 0 neutron.
So net charge on each hydrogen atom
=Change on 1 electron + charge on 1 proton
=(-e)+(e+Δe)
=Δe
According the Question
Electrostatic force(Repulsion) + Graviational force (Attraction) = 0
Magnitude of Electrostatic Force = Magnitude of Gravitational Force
kq1q2/d2 = Gm1m2/d2
k(Δe)2 = G(mh)2
(Δe)2=G(mh)2/k
Δe = √[G(mh)2/k]
Δe = √[(6.67×10-11)(1.67×10-27)2/(9×109)]
Δe = √[(6.67×1.67×1.67×10-74)/9]
Δe = √[2.06688×10-74]
Δe = √[2.06688×10-74]
Δe = 1.4376×10-37
Δe ≈ 1.4×10-37C
Hence, c) 10-37C would be the correct option.
