Question

Compute the gravitational force and the electric force between the electron and the proton in the hydrogen atom if they are 5.3 x 10-11 meters apart. Then calculate the ratio of Fe to Fg.

Answer 1

Answer:

Fg = 3.61×10⁻⁴⁷ N.

Fe = 8.2× 10⁻⁸ N.

Fe/Fg = 2.27×10³⁹

Explanation:

From Newton' s law,

Fg = Gm'm/r²...................... Equation 1

Where Fg = gravitational force, m = mass of proton, m' = mass of electron, r = distance between proton and electron, G = Gravitational constant

Given:  m = 1.67 x 10⁻²⁷ kg, m' =  9.1×10⁻³¹ kg, r = 5.3×10⁻¹¹ m

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 1

Fg = (1.67 x 10⁻²⁷)( 9.1×10⁻³¹ )(6.67×10⁻¹¹)/( 5.3×10⁻¹¹)²

Fg = (101.364×10⁻⁶⁹)/(28.09×10⁻²²)

Fg = 3.61×10⁻⁴⁷ N.

From coulomb's law,

Fe = kqq'/r²................. Equation 2

Where Fe = Electric force, q = charge of proton, q' = charge of electron, r = distance between proton and electron.

Given: q = 1.6×10⁻¹⁹ C, q' = 1.6×10⁻¹⁹ C, r = 5.3×10⁻¹¹ m, k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

Fe = (1.6×10⁻¹⁹)(1.6×10⁻¹⁹)(9.0×10⁹)/(5.3×10⁻¹¹)²

Fe = 23.04×10⁻²⁹/(28.09×10⁻²²)

Fe = 8.2× 10⁻⁸ N.

Therefore,

Fe/Fg = 8.2× 10⁻⁸ /3.61×10⁻⁴⁷

Fe/Fg = 2.27×10³⁹

Answer 2

Fg = 3.61×10⁻⁴⁷ N.

Fe = 8.2× 10⁻⁸ N.

Fe/Fg = 2.27×10³⁹

Explanation:

From Newton's law of gravitational force ofattraction,

Gravitational force of attraction, Fg = Gm1m2/r²

Where,

m1 = mass of electron

= 9.1 x 10⁻³¹ kg

m2 = mass of protron

= 1.67 x 10⁻²⁷ kg

r = distance between proton and electron

= 5.3×10⁻¹¹ m

G = Gravitational constant

= 6.67×10⁻¹¹ Nm²/kg²

Therefore,

Fg = (1.67 x 10⁻²⁷ × 9.1×10⁻³¹ × 6.67×10⁻¹¹)/( 5.3×10⁻¹¹)²

= 3.6085 × 10⁻⁴⁷ N.

From coulomb's law,

Electric force, Fe = kq1q2/r²

Where,

q1 = charge of proton

= 1.6×10⁻¹⁹ C

q2 = charge of electron

= 1.6×10⁻¹⁹ C

r = distance between proton and electron.

= 5.3×10⁻¹¹ m

k = 9.0×10⁹ Nm²/C²

Fe = (1.6×10⁻¹⁹ × 1.6×10⁻¹⁹ × 9.0×10⁹)/(5.3×10⁻¹¹)²

= 8.202 × 10⁻⁸ N.

Therefore, ratio of Fe to Fg:

Fe/Fg = 8.202 × 10⁻⁸/3.6085 × 10⁻⁴⁷

= 2.273 × 10³⁹