An early planetary model of the hydrogen atom consisted of a 1.67 X 10-27 kg proton in the nucleus and a 9.11 X 10-31 kg electron in orbit around it at a distance of 5.0 X 10-11 m. In this model, what is the gravitational force between a proton and an electron?

Respuesta :

Answer:

The gravitational force between proton and an electron is [tex]4.06 \times 10^{-47}[/tex] N

Explanation:

Given:

Mass of proton [tex]m_{1} = 1.67 \times 10^{-27}[/tex] kg

Mass of electron [tex]m_{2} = 9.11 \times 10^{-31}[/tex] kg

Separation between electron and proton [tex]r = 5 \times 10^{-11}[/tex] m

According to the gravitational law,

   [tex]F = \frac{Gm_{1} m_{2} }{r^{2} }[/tex]

Where [tex]G = 6.674 \times 10^{-11}[/tex] = gravitational constant.

   [tex]F = \frac{6.674 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31} }{(5 \times 10^{-11} )^{2} }[/tex]

   [tex]F = 4.06 \times 10^{-47}[/tex] N

Therefore, the gravitational force between proton and an electron is [tex]4.06 \times 10^{-47}[/tex] N

The gravitational force between a proton and an electron is 40 * 10⁻⁴⁸ N

Newton's law of gravitation is given by:

F = Gm₁m₂/r²

Where F is the force, m₁, m₂ are masses, r is the distance between the two masses and G is the gravitational constant = 6.67 * 10⁻¹¹ Nm²/kg²

Given that:

m₁ = 1.67 * 10⁻²⁷ kg, m₂ = 9.11 * 10⁻³¹ kg, r = 5 * 10⁻¹¹ m, hence:

F = 6.67 * 10⁻¹¹ Nm²/kg² * 1.67 * 10⁻²⁷ kg * 9.11 * 10⁻³¹ kg/ (5 * 10⁻¹¹ m)²

F = 40 * 10⁻⁴⁸ N

The gravitational force between a proton and an electron is 40 * 10⁻⁴⁸ N

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