Hydrogen transition probability

In summary, I solved the integral for an arbitrary m and l=1, calling the result of the <100|r|21m>=\gamma. Since it does not depend on t we can take it out of the integral along with all the constants. The integral is then:P = \gamma^2 \epsilon^2 \left| \int_0^{T \gg \tau} e^{-\left( \frac{3i}{4} + \frac{\alpha}{\tau}\right)t} dt \right|^2For which I solved and took the limit T \rightarrow \infty and got the result:\frac{\
  • #1
svletana
21
1
Hello! I have the following problem I'm trying to solve:

Homework Statement


An Hydrogen atom in the state |100> is found between the plates of a capacitor, where the electric field (weak and uniform) is: [itex]E(t) = \epsilon e^{-\alpha t / \tau}[/itex].

Calculate the parameters of the potential ([itex]\epsilon, \alpha, \tau[/itex]) so that for a time [itex]t \gg \tau[/itex] the transition probability to any of the n=2 states is equal to 0.1.

Homework Equations


The field is asumed to be in an arbitrary [itex]r = (x,y,z)[/itex] direction, so that [itex]W = \epsilon e^{-\alpha t / \tau} r[/itex].

The formula for transition probability is (using atomic units):
[tex]P = \left| \int_0^{T \gg \tau} e^{i \omega t} <100 | r | 21m> \epsilon e^{-\alpha t / \tau}dt \right|^2[/tex]

where [itex]\omega = \frac{E_{21m} - E_{100}}{\hbar} = \frac{-3}{4} [/itex].

For the [itex]<100|r|21m>[/itex] elements we have the results, for each m:
[tex]<100|r|200> = 0[/tex]
[tex]<100|r|210> = \frac{2^7 \sqrt{2} \hat{z}}{3^5}[/tex]
[tex]<100|r|21\pm1> = \frac{2^7}{3^5}(\mp\hat{x} - i\hat{y})[/tex]

The Attempt at a Solution


I solved the integral for an arbitrary m and l=1, calling the result of the [itex]<100|r|21m>=\gamma[/itex]. Since it does not depend on t we can take it out of the integral along with all the constants. The integral is then:

[tex]P = \gamma^2 \epsilon^2 \left| \int_0^{T \gg \tau} e^{-\left( \frac{3i}{4} + \frac{\alpha}{\tau}\right)t} dt \right|^2[/tex]

For which I solved and took the limit [itex]T \rightarrow \infty[/itex] and got the result:

[tex]\frac{\gamma^2 \epsilon^2}{\frac{9}{16} + \frac{\alpha^2}{\tau^2}}[/tex]

Now, [itex]\gamma^2 = \frac{2^{15}}{3^{10}}[/itex] for any value of m. And that result must be equal to 0.1 according to the guidelines.

I don't see how I could possibly calculate 3 parameters from this equation, what am I missing? Thanks in advance!
 
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  • #2
Not that this is that much help but shouldn't the potential be proportional to [itex] r \cos \theta [/itex] not [itex] r[/itex]? With this you get zero for [itex]m=-1[/itex], but the same otherwise.
 
  • #3
Dazed&Confused said:
Not that this is that much help but shouldn't the potential be proportional to [itex] r \cos \theta [/itex] not [itex] r[/itex]? With this you get zero for [itex]m=-1[/itex], but the same otherwise.

Sorry, r would be a vector, for example r = xî + yĵ + zk, so you could have it in any of the three directions.
 
  • #4
Ah ok. It's just I would have thought you take the inner product of the potential, that is, the potential is your operator.
 
  • #5
I just talked to my teacher and he said that i should choose the field going in the z direction so the only transition possible is to the |210> state and then choose parameters that would fit the problem... Guess that settles it :P
 
  • #6
But I'm also not sure how you can get 3 parameters from one equation. It is strange that it says for any n=2 state though, when at least if l=1, m=0, you got 0 as the probability. Also perhaps that the electric field is weak needs to be employed.
 
  • #7
Well that's what I did, hence [itex] z = r \cos \theta [/itex], however to me that didn't give an obvious solution.
 
  • #8
Would the parameters [itex]\alpha = 3.67 \times 10^7[/itex], [itex]\epsilon = 1.93 \times 10^{11} \frac{V}{m}[/itex] and [itex]\tau = 1.109 \times 10^{-9} [/itex] seconds work?

I had in mind that it would have to be an electric field strong enough to make the transition but not so big so hydrogen is ionized, a value of [itex]\tau[/itex] around the value of the half life for that transition, and calculating [itex]\alpha[/itex] with those two and the expression.
 
  • #9
That sounds reasonable. The wording of question doesn't imply you have some freedom in choosing your parameters, but since the number of equations is under determined I would say you are likely right.
 
  • #10
Yeah I only really found out because I emailed my teacher asking how would I get 3 parameters from one equation, but that's exactly what he wrote...

Thanks for your help! :)
 

What is hydrogen transition probability?

Hydrogen transition probability refers to the likelihood that an electron in a hydrogen atom will change energy levels or states. It is used to describe the probability of absorption or emission of photons by hydrogen atoms.

How is hydrogen transition probability calculated?

Hydrogen transition probability is calculated using the transition probability formula, which takes into account the energy levels of the electron before and after the transition, as well as the quantum numbers associated with those levels.

What is the significance of hydrogen transition probability in physics?

Hydrogen transition probability plays a crucial role in understanding the behavior of atoms and molecules. It helps us to predict and explain the absorption and emission spectra of hydrogen, which is important in fields such as astrophysics and spectroscopy.

Can hydrogen transition probability be observed experimentally?

Yes, hydrogen transition probability can be observed experimentally through spectroscopy. By measuring the wavelengths of photons emitted or absorbed by hydrogen atoms, we can determine the probabilities of different transitions between energy levels.

How does hydrogen transition probability relate to the uncertainty principle?

Hydrogen transition probability is related to the uncertainty principle in that it represents the probability of an electron transitioning between energy levels, rather than a definite position or energy. This is consistent with the uncertainty principle, which states that we cannot simultaneously know the exact position and momentum of a particle.

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