Question

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm. (a) Find the magnitude of the electrostatic force of attraction, Fe between the electron and the proton. (b) Find the magnitude of the gravitational force of attraction Fg , between the electron and the proton, and find the ratio, Fe /Fg . me = 9.11 x 10-31kg, e = 1.602 x 10-19C mp = 1.67 x 10-27kg k = 9 x 109 Nm2 /C2 G = 6.67 x 10-11 Nm2 /kg2

Answer 1

Answer:

(a)  8.2 x 10^-8 N

(b) 3.6 x 10^-47 N  , 2.27 x 10^39

Explanation:

charge of proton, q1 = 1.6 x 10^-19 C

charge of electron, q2 = - 1.6 x 10^-19 C

radius of orbit, r = 0.053 nm = 0.053 x 10^-9 m

mass of electron, me = 9.1 x 10^-31 kg

mass of proton, mp = 1.67 x 10^-27 kg

Gravitational constant, G = 6.67 x 10^-11 Nm^2/kg^2

(a) The electrostatic force between two charges is given by

$F_{e}=\frac{Kq_{1}q_{2}}{r^2}$

Where, K is the coulombic constant = 9 x 10^9 Nm^2/C^2

By substituting the values

$F_{e}=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\1.6\times 10^{-19}}{\left ( 0.053 \times 10^{-9} \right )^2}$

Fe = 8.2 x 10^-8 N

(b) The gravitational force between the electron and proton is given by

$F_{g}=\frac{Gm_{e}m_{p}}{r^{2}}$

$F_{g}=\frac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\1.67\times 10^{-27}}{\left ( 0.053 \times 10^{-9} \right )^2}$

Fg = 3.6 x 10^-47 N

$\frac{F_{e}}{F_{g}}=\frac{8.2\times10^{-8}}{3.6\times 10^{-47}}$

$\frac{F_{e}}{F_{g}}=2.27\times 10^{39}$