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 of 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 [Givenmass of hydrogen mh = 1.67 × 10⁻²⁷ kg](a) 10⁻²³ C(b) 10⁻³⁷ C(c) 10⁻⁴⁷ C(d) 10⁻²⁰ C

Answer 1

answer : option (b) 10^-37 C

explanation : we know, a hydrogen atom has one electron and one proton

so, net charge on each hydrogen atom = (e + ∆e - e) = ∆e

so, electrostatic force , $F_e=\frac{K\Delta e^2}{d^2}$

gravitational force, $F_g=\frac{GM_hM_h}{d^2}$

where $M_h$ denotes mass of hydrogen

a/c to question,

gravitational force = electrostatic force

so, $\frac{K\Delta e^2}{d^2}=\frac{GM_h^2}{d^2}$

or, $K\Delta e^2=GM_h^2$

we know, K = 9 × 10^9 Nm²/C² , G = 6.67 × 10^-11 Nm²/Kg² , $M_h$ = 1.67 × 10^-27 Kg

so, 9 × 10^9 × ∆e² = 6.67 × 10^-11 × (1.67 × 10^-27)²

or, ∆e² = 6.67 × 10^-11 × 1.67 × 1.67 × 10^-54/9 × 10^9

or, ∆e = 1.43767 × 10^-37 C

hence, option (b) is correct.

Answer 2

Answer:

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