Derivation of orbital period - Hydrogen

In summary, to derive the formula for T_{Kepler} from the given wave packet, we use a first order Taylor series approximation of the phase with respect to the variable n. This results in the equation T_{Kepler}=2\pi \bar n ^3, which does not have the dimension of seconds.
  • #1
AwesomeTrains
116
3
Hey everyone!
1. Homework Statement

I've been giving the equation for a gaussian wave packet and from that I have to derive this formula:
[itex]T_{Kepler}=2\pi \bar n ^3[/itex] by doing a first order taylor series approximation at [itex]\bar n[/itex] of the phase:
[itex]f(x)=f(\bar n)+\frac{df}{dx}|_{\bar n}(x-\bar n)+\sigma(x^2)[/itex]

The given wave packet:
[itex]\psi(t)=\frac{1}{(2\pi\sigma_n^2)^{1/4}}\sum_{n=1}^\infty exp\left(-\frac{(n-\bar n)^2}{4\sigma_n^2}\right)r\psi_{n,l,m}exp\left(i\frac{t}{2n^2}\right)[/itex]

Homework Equations


I assume I've posted all the relevant ones in 1.

The Attempt at a Solution


Why is the dimension of [itex]T_{Kepler}[/itex] not seconds?
Anyways I tried doing the taylor approximation with t as the differentiation variable and got this (assuming [itex]i\frac{t}{2n^2}[/itex] is the phase):
[itex]i\frac{t}{2n^2}=i\frac{\bar n}{2n^2}+i\frac{t-\bar n}{2n^2}+\sigma(t^2)=i\frac{it}{2n^2}[/itex] (Strange)

Am I supposed to do it for the variable n? Or how do I proceed from here?
Any hints are very appreciated.
Kind regards
Alex
 
Last edited:
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  • #2
2. The equation you are trying to approximate is for the phase of the wave packet, so you should differentiate with respect to the variable n. You can use the chain rule to break up the derivative into terms that are easier to evaluate.Start by writing the phase as a function of n:f(n)=f(\bar n)+\frac{df}{dn}|_{\bar n}(n-\bar n)+\sigma(n^2)Now take the derivative with respect to n: \frac{df}{dn}=\frac{df}{dn}|_{\bar n}+2\sigma nEvaluate this at \bar n and substitute it back into your equation:f(n)=f(\bar n)+\frac{df}{dn}|_{\bar n}(n-\bar n)+\sigma(n^2)=f(\bar n)+2\sigma \bar n (n-\bar n)+\sigma(n^2)Now do a first order Taylor series approximation, meaning you only keep terms up to the first power of n-\bar n:f(n)\approx f(\bar n)+2\sigma \bar n (n-\bar n)Finally, solve for T_{Kepler}:T_{Kepler}=2\pi \bar n ^3
 

Related to Derivation of orbital period - Hydrogen

1. What is the equation for calculating the orbital period of a hydrogen atom?

The equation for calculating the orbital period of a hydrogen atom is T = 2π√(a^3/GM), where T is the orbital period, a is the semi-major axis of the orbit, G is the gravitational constant, and M is the mass of the central body (in this case, the nucleus of the hydrogen atom).

2. How is the orbital period of a hydrogen atom derived?

The orbital period of a hydrogen atom is derived using Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

3. Why is the mass of the central body used in the equation for orbital period of a hydrogen atom?

The mass of the central body (M) is used in the equation because it determines the strength of the gravitational force between the hydrogen atom and the nucleus. This force is what keeps the atom in orbit, and therefore, it is essential in calculating the orbital period.

4. Does the orbital period of a hydrogen atom change at different distances from the nucleus?

Yes, the orbital period of a hydrogen atom does change at different distances from the nucleus. According to Kepler's third law, the orbital period is directly proportional to the semi-major axis of the orbit. Therefore, the farther the hydrogen atom is from the nucleus, the longer its orbital period will be.

5. How does the orbital period of a hydrogen atom compare to other elements?

The orbital period of a hydrogen atom is relatively short compared to other elements, as it only has one proton in its nucleus. Elements with larger nuclei and more electrons will have longer orbital periods due to the increased strength of the gravitational force between the nucleus and the electrons.

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