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- #2

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Can you show the math?

- #3

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v = e

Using m=m

and E=(m-m

Using Bohr's equations E = m*e

I tried to see what can be derived by substituting the first equation in the second by get stuck in the math.

- #4

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One other note: the relativistic correction to the mass of the electron due to its zipping around the nucleus will only be like 3 parts in 100,000, so you won't notice a difference if you calculate the energy to just the first decimal place.

What the Bohr model does:

Predicts the energy levels of the hydrogen atom to a first-order approximation (the coulomb potential, non-relativistic kinetic energy, and nothing else):

What the Bohr model does not include:

Relativistic corrections to the kinetic energy

Corrections due to the magnetic field created by the orbiting electron interacting with its own magnetic moment (Spin-orbit coupling)

Corrections due to the nucleus having a nonzero size (Darwin term)

Corrections due to the interactions of the electron's and proton's magnetic dipoles (spin-spin coupling).

among others...

Also, even without these corrections, the Bohr model doesn't predict what the electron cloud should look like for a hydrogen atom in a given state, or what the likelihood is for an atom to decay from an excited state to the ground state in a given time.

If you want a good model of the Hydrogen atom, you're going to want to solve the Schrodinger equation describing the electron-proton interaction. You'll get the same results as the Bohr model for the first-order approximation, but it's easier to get theoretical predictions that more closely agree with experimental results by including these extra terms.

- #5

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But that's what surprises me so much. The energy difference due to this relativistic effect of "only" 3 in 100,000 (to be more precise 2.66 in 100,000) is the same as the energy level of the electron from the Bohr model:One other note: the relativistic correction to the mass of the electron due to its zipping around the nucleus will only be like 3 parts in 100,000, so you won't notice a difference if you calculate the energy to just the first decimal place.

2.66/100000 * mc

- #6

blue_leaf77

Science Advisor

Homework Helper

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$$

mc^2 \approx m_0 \left( 1+\frac{v_n^2}{2c^2} \right) c^2

$$

thus

$$

E = (m-m_0)c^2 \approx \frac{m_0 v_n^2}{2} = \frac{m_0 e^4 Z^2}{8\epsilon_0^2 h^2 n^2}

$$

which is,

- #7

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I simply calculated the kinetic energy since v<<c. I forgot to add the potential energy, which is twice the kinetic energy but negative, hence the sign error in my result.

- #8

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As was already mentioned, the relativistic formula for KE is not necessary for this case, you could use the classic one.

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