Suppose the charge of a proton and an electron differ slightly. One of them is -e and the other is ( e+ e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed…

Question

Suppose the charge of a proton and an electron differ slightly. One of them is -e and the other is ( e+ 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 e is of the order [Given mass of hydrogen, m_h=1.67 10^-27 kg] [NEET 2017]

Accepted Answer

Correct answer: C. (c) Net charge on one H -atom q=-e+e+ e= e Net electrostatic repulsive force between two H -atoms F_r= Kq^2d^2= K ( e )^2d^2 Similarly, net gravitational attractive force between two H-atoms, F_G= Gm_r^2d^2 It is given that, F_r-F_G=0 ⇒ K ( e )^2d^2- Gm_r^2d^2=0 ⇒ ( e )^2= Gm_r^2K . ( e )^2= (6.67 10^-11)(1.67 10^-27)^29 10^9 ⇒ e=1.437 10^-37C

Solution

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