Absorption prob - why my way isn't correct?

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In summary, the conversation discussed the energy difference between two levels of a hydrogen electron. The correct method for finding this difference is by using the energy of the emitted photon, rather than trying to subtract the energies of the two levels. This can be seen by drawing a diagram of the energy levels and transitions. The energy difference was found to be 1.55*10^-19 J.
  • #1
assaftolko
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Homework Statement


A hydrogen electron is at its ground level when it's radiated in wave length of 94.92 nm. As a result it arrises to level a. then emission occures with frequency of 2.34*10^14 Hz and the electron falls to level b. What's the energy difference between the 2 levels

Homework Equations


E=hv, c=λv

The Attempt at a Solution


I found E1 using λ1 that's given to me, E2 using v2 given to me and substracted E2 from E1. Why isn't this way correct and you have to use rydberg's formula? whay the 2 don't give the same result? I got 1.94*10^-18 J
 
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  • #2
You don't have to use Rydberg's formula, but is
assaftolko said:
I found [...] E2 using v2
correct?
 
  • #3
Why not? Thought so...
 
  • #4
assaftolko said:
Why not? Thought so...

Make a diagram with the energy levels and the transitions, you will see.
 
  • #5
Im sorry i don't quite get how to understand my mistake...
 
  • #6
Ok, let's try another approach. Can you tell me what happens in the last step of the process?
 
  • #7
DrClaude said:
Ok, let's try another approach. Can you tell me what happens in the last step of the process?

Energy is emitted from the electron so it "drops" to a lower energy level, closer to the ground level than it was at first after it arrived to level a
 
  • #8
And to what does the energy of the emitted photon correspond?
 
  • #9
DrClaude said:
And to what does the energy of the emitted photon correspond?

to hv... so what you're saying is that this energy is not the energy of level b with respect to the ground level, but with respect to level a?
 
  • #10
Exactly! You get directly the the difference in energy between the levels from the emitted photon. The difference between the ground state and level a is actually irrelevant here.
 
  • #11
DrClaude said:
Exactly! You get directly the the difference in energy between the levels from the emitted photon. The difference between the ground state and level a is actually irrelevant here.

that's weird because you still don't get the correct answer - which should be 1.55*10^-19 J

Oh **** yes it does... Oh man :/

Thanks!
 
  • #12
You're welcome!
 

Related to Absorption prob - why my way isn't correct?

1. Why is absorption probability important in scientific research?

Absorption probability is important because it helps us understand how much of a substance or energy is taken up by a material. This is crucial in many fields of research, such as medicine, environmental science, and material science, to name a few.

2. How is absorption probability calculated?

The absorption probability of a material can be calculated by dividing the amount of the absorbed substance or energy by the total amount of the substance or energy present. This is commonly expressed as a percentage.

3. Why might my approach to calculating absorption probability be incorrect?

Your approach may be incorrect if you are not taking into account all the factors that can affect absorption, such as the concentration of the substance, the thickness of the material, and the wavelength of the energy. It is important to consider all these factors in order to accurately calculate absorption probability.

4. How can I improve the accuracy of my absorption probability calculations?

To improve the accuracy of your absorption probability calculations, it is important to carefully measure and control all the factors that can affect absorption, as mentioned earlier. It is also helpful to consult with other scientists in the field and use established methods and techniques for calculating absorption probability.

5. What are some real-world applications of absorption probability?

Absorption probability has many real-world applications, including drug delivery in medicine, determining the effectiveness of air filters in environmental science, and designing solar panels in material science. It is also used in spectroscopy to analyze the composition of substances and in nuclear physics to study the interactions of particles with matter.

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