calculate the coulombos force b/w electron and nucleus in normal hydrogen atom?
Answers
Answer:
Tricky question.
If you pretend that the electron is a perfectly localized speck orbiting the proton like the Earth orbits the Sun, you can calculate the electrostatic force from Coulomb’s law and from that relate the energy to the radius, etc.
But the electron is not a localized speck. If it were, and if that picture were correct, it would radiate away all its energy in EM waves (light) and fall into the proton within about a nanosecond and all matter would be unstable.
In reality the electron is in a wavefunction with different probabilities of being at different distances from the proton, at each of which the Coulomb force would be different. One can calculate the average force, or the expectation value of the force, but that’s not really what you asked for.
Answer:
This problem involves repeated application of Coulomb's law of electric forces and Newton's law of universal gravitation.
Coulomb's law for the hydrogen atom…
FE =
kq1q2
r2
FE =
(9.0 × 109 N m2/C2)(1.60 × 10−19 C)2
(0.053 × 10−9 m)2
FE = 8.2 × 10−8 N
Newton's law for the hydrogen atom…
Fg =
Gm1m2
r2
Fg =
(6.67 × 10−11 N m2/kg2)(1.67 × 10−27 kg)(9.11 × 10−31 kg)
(0.053 × 10−9 m)2
Fg = 3.6 × 10−47 N
The electric force may seem like a small number, but keep in mind that the electron doesn't have much mass. The electric force is sufficient to keep the electron in "orbit" around the proton in the hydrogen atom. At 39 orders of magnitude smaller, the gravitational force might as well be zero. Gravity does not do anything to keep a hydrogen atom together.
Compare…
FE =
8.2 × 10−8 N
3.6 × 10−47 N
Fg
FE ≈ 1039 or 39 orders of magnitude
Fg
Gravity is a weak force on the atomic scale.