Question

The distance between the electron and proton in hydrogen atom is 5.3×10 to the power minus 11m ,caculate the electrostatic force of attraction between them
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Answer 1

distance between electron and proton in a hydrogen atom is 5.3 × 10^-11 m

charge on electron , e = -1.6 × 10^-19C

charge on proton, p = +1.6 × 10^-19C

now from Coulomb's law,

F = K(e)(p)/r²

here, r = 5.3 × 10^-11 m and K = 9 × 10^9 Nm²/C²

so, F = 9 × 10^9 × (-1.6 × 10^-19)(+1.6 × 10^-19)/(5.3 × 10^-11)²

= -9 × 2.56 × 10^-29/(5.3 × 5.3 × 10^-22)

= -0.82 × 10^-7

= -8.2 × 10^-8 N

[here negative sign indicates that force is attractive]

hence, magnitude electrostatic force of attraction between electron and proton is 8.2 × 10^-8 N

Answer 2

Explanation:

charge on the electron and proton is same i.e. 1.6×10^-19 C. and. distance is 5.33× 10^-11 m

by applying coulombs law we get the force between the electron and proton is 8.1×10^-8 N