Question

In hydrogen atom the distance between nucleus and electron is 53pm .find force between nucleus and electron.

Answer 1

Explanation:

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