O Physics: Why doesn’t the electron crash into the proton?

[jtbell]

Does this mean that when a proton and a neutron bind together as a nucleus, that leaves them in a lower energy state, so energy is released and the result is that the nucleus has less mass than the proton and neutron? And is that the source of the energy that is released in fusion?

You got it!

And conversely, in order to separate the proton + neutron (a deuterium nucleus) you have to do work on them, which increases the total mass.

I see, if I add up the masses of 1 tritium atom at 3.016 u plus 1 hydrogen atom at 1.008 (total 4.024); or 2 deuterium atoms (total 4.028 u) whereas the mass of a helium atom is 4.003 u.

Is that difference of about 0.02 u per atom of helium where the energy released by a hydrogen bomb comes from?

Right. In doing calculations like this, sometimes you have to take into account the difference between atomic masses (which is what tables show) and nuclear masses, but that's not the case here since you have the same number of atomic electrons before and after. In principle, the slight difference in the binding energies of the atomic electrons to the nuclei might also make a difference, but it's so tiny (a few eV compared to millions of eV) that it's not significant for most purposes.

 

[Andrew Mason]

?
well I'm lost

i presume the the Earth doesn't crash into the sun because of the speed at which it orbits, is the reason that it orbits the convection currents on the sun?

so simply, does or doesn't the electron orbiting the proton act like the Earth orbiting the sun?

as an awnser to the guy saying about Earth not being able to lose energy because of the lack of atmosphere in space, well what about reflecting light and heat...

finally, if the sun gives off light and heat energy to earth, then would Earth never lose enough energy to lose orbit and crush into the sun, atleast before the sun loses all H anyway

This raises a very interesting question. Using classical mechanics and electomagnetism, work out the speed that an electron would have to have in order to orbit a hydrogen nucleus at a distance of 10^-12 m:

$$F_c = \frac{m_ev^2}{r} = \frac{kq_e^2}{r^2}$$

$$v = \sqrt{\frac{kq_e^2}{mr}}$$

where:
$r = radius of orbit = 1e-12 m$
$k = 9e9 Nm^2/C^2$
$q_e = 1.602e-19 C.$
$m_e = 9.1e-31 kg$

v works out to 1.6e7 m/sec or about 5% of the speed of light.

Now work out what the radius of orbit could be if the electron traveled at the speed of light. This would obviously be the minimum orbital radius permitted by relativity. It would take an infinite amount of energy for an electron to get arbitrarily close to the speed of light.

I get r = 2.5e-15 m. or 2.5 Fermi units

The radius of a proton is about .5 Fermi. To reach a 2.5 Fermi radius of orbit, the electron would need an infinite amount of energy. So an orbiting electron simply can't get enough energy to crash into the nucleus!


I just made this up, so I might be missing something here. But it does seem to explain why there is a limit to how close orbiting electrons and protons can come.

AM

 

[Antimatter]

For about 92 years it has been scientifically proven that electrons DO NOT behave like (very little) planets.

Danie

Indeed, the elektron would loose energy because it would radiate (being an accelerated charged particle & all); atom collapse is predicted in the order of 10^(-10) seconds if I recall correctly.

 

[Andrew Mason]

Indeed, the elektron would loose energy because it would radiate (being an accelerated charged particle & all); atom collapse is predicted in the order of 10^(-10) seconds if I recall correctly.

That is only if the theory that an electron radiates because it accelerates is true. It is still somewhat controversial and not proven. See a recent discussion of this at:
https://www.physicsforums.com/showthread.php?t=65767

AM