Quantum physics hydrogen atom releasing a photon?

I think the key might lie in the fact that if the speed of a photon is the speed of light then the photon must be regarded as an ultrarelatavistic particle.

so p = E/c

The energy of a particle with a speed equal to the speed of light can be finite only if the mass is zero:

so photons must be regarded as particles of zero mass.

Since the energy of a photon is E = hf = hc / wavelength

Momentum is therefore p = h / wavelength

My quantum mechanics knowledge is not of a high standard, but if the mass is zero where does the interaction occur with the bulk of the atom, and does stopping potential enter into the problem.

Another possible item for thought would be a wavelength change between absorption and emission when a photon collision occurs.

Since the kinetic energy of a hydrogen atom in any given state is ~= -13.606 eV * ( 1 / n^2) we know the greatest change in energy will occur between the first excited state (n = 2) and the ground state (n = 1).

This change can be calculated as E = -13.606 * ( 1 - (1 / 2^2) ) = -10.2045 eV

Changing energy levels will release a photon to compensate for the energy loss of the hydrogen atom, and the photon will have kinetic energy exactly equal to the change in kinetic energy of the hydrogen atom. Since the kinetic energy of a photon can be expressed as the product of its momentum times the speed of light:

KE = pc

we can say that the momentum of the particle is equal to the kinetic energy divided by the speed of light

p = KE / c

However, before performing this calculation the kinetic energy of the photon should be converted into joules. This needs to be done anytime you are doing a calculation that involves a particle with mass (in this case the hydrogen atom.

1eV = 1.602 E-19 J

10.2045 eV = 1.63476 E-18 J

Now we can calculate momentum:

p = 1.63476 E-18 J / 2.998 E8 * (m/s) = 5.45283 E-27 kg * (m/s)

Since the total momentum of this hydrogen - photon system must be conserved we know that the hydrogen that was once at rest, meaning p = 0, must gain an equivalent (however opposite) momentum to compensate for the photon. So we set to momentum of the photon equal to the momentum of the hydrogen atom. Since the momentum of a particle with mass is given by mass * velocity, this can be used to easily calculate the recoil velocity of the hydrogen.

5.45283 E-27 kg * (m/s) = mass of hydrogen * recoil velocity {mass of H = 1.674 E-27)

recoil velocity = 3.26 m/s

Note: Energy and momentum of massless particles such as photons are calculated differently than particles with mass. Photons do have energy, which in turn means they must also have momentum. This is why the atom will "recoil" or in other words change velocity when it releases a photon.

at first. A "H" is two plus 2. text textile e book de filne, no such subject as "relax" . 2d of all, no way will an "H" emits photons. Bang, it is that. Lessin, you obtain a great colider on your basement" . And 0.33, i'm existence.