[DESPERATE] Frank-Hertz/Compton/De Broglie

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In summary, the conversation discusses questions related to an experiment involving an element X which produces an emission spectrum with specific wavelengths. The questions involve determining kinetic energy, verifying Bohr's model of the atom, calculating momentum, and predicting the behavior of light waves after colliding with different particles. The conversation also includes a new question about calculating the wavelength of an electron in the 4th energy level of a hydrogen atom, and asks for help in understanding Compton Scattering.
  • #1
alexlee33
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If you can help me on even one of these questions, it's greatly appreciated!

Homework Statement



Use the following info to answer next 4 questions:

The vapor of Element X produces an emission spectrum that consists of wavelengths of 400 nm, 550 nm, and 680 nm.

1) If an unexcited sample of element x is bombarded w/ elections that have a kinetic energy of 2.00 eV, then determine the amnt of kinetic energy the e- wouild have as the exit the sample.

2) Explain how this observation verifies Bohr's model of the atom.

3) Determine the momentum of an emitted 570 nm light wave.

4) Decribe what would happen to the direction and wavelength of a 570 nm light wave after it collides with:
a) the nucleus of a large atom
b) an electron

------- new Q -------

5) Calculate the wavelength of an e- in the 4th energy level of a hydrogen atom.

Homework Equations



See below.
And also E2 = E1 / n2
E1 = -2.18 x 10^-18 OR -13.6 eV

E = hc / (wavelength)
h= 6.63 x 10^-34
There may be more; I'm not sure which to use! Please help!

The Attempt at a Solution



1) This is one I have no idea... I've tried different things but keep getting the wrong answer, I believe.
I tried finding E2, and I assumed the energy level # is 2 because it's in the visible light spectrum.
So I used E2 = E1/n2, and then I subtracted E1 from E2, which gives me the answer to how much energy the atom can absorb. Then I tried to subtract that answer from the enegry given to me in the question, but I'm in doubt that this is correct.

2) Bohr states that an atom is a nucleus surrounded by e- moving in circular orbits and that energy levels are quantized. This expt (Frank-Hertz) proves that e- truly receive energy in discrete amounts, and energy need is based on energy levels.

3) p = h / (wavelength)
p = (6.63 x 10^-34) / (570 x 10^-9)
FINAL ANSWER: p = 1.16 x 10^-27 kg(m/s)

4 a) Bounce back?? What happens to the wavelength? I'm stumped.
4 b) I have no idea.

5) En = E1/n^2
En = -2.18 x 10^-18 / 4^2 = -1.3325 x 10^-19
E = hc / (wavelength)
(wavelength) = hc / E = ((6.63 x 10^-34)(3x10^8)) / (-1.3625 x 10^-19)
FINAL ANSWER: -1.46 x 10^-6
 
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  • #2
alexlee33 said:
If you can help me on even one of these questions, it's greatly appreciated!

Homework Statement



Use the following info to answer next 4 questions:

The vapor of Element X produces an emission spectrum that consists of wavelengths of 400 nm, 550 nm, and 680 nm.

1) If an unexcited sample of element x is bombarded w/ elections that have a kinetic energy of 2.00 eV, then determine the amnt of kinetic energy the e- wouild have as the exit the sample.

2) Explain how this observation verifies Bohr's model of the atom.

3) Determine the momentum of an emitted 570 nm light wave.

4) Decribe what would happen to the direction and wavelength of a 570 nm light wave after it collides with:
a) the nucleus of a large atom
b) an electron

------- new Q -------

5) Calculate the wavelength of an e- in the 4th energy level of a hydrogen atom.

Homework Equations



See below.
And also E2 = E1 / n2
E1 = -2.18 x 10^-18 OR -13.6 eV

E = hc / (wavelength)
h= 6.63 x 10^-34
There may be more; I'm not sure which to use! Please help!

The Attempt at a Solution



1) This is one I have no idea... I've tried different things but keep getting the wrong answer, I believe.
I tried finding E2, and I assumed the energy level # is 2 because it's in the visible light spectrum.
So I used E2 = E1/n2, and then I subtracted E1 from E2, which gives me the answer to how much energy the atom can absorb. Then I tried to subtract that answer from the enegry given to me in the question, but I'm in doubt that this is correct.
Try converting each of the wavelengths into its respective energy.
alexlee33 said:
3) p = h / (wavelength)
p = (6.63 x 10^-34) / (570 x 10^-9)
FINAL ANSWER: p = 1.16 x 10^-27 kg(m/s)
Correct.
alexlee33 said:
4 a) Bounce back?? What happens to the wavelength? I'm stumped.
4 b) I have no idea.
What does Compton Scattering tell you? (look at the equation).
alexlee33 said:
5) En = E1/n^2
En = -2.18 x 10^-18 / 4^2 = -1.3325 x 10^-19
E = hc / (wavelength)
(wavelength) = hc / E = ((6.63 x 10^-34)(3x10^8)) / (-1.3625 x 10^-19)
FINAL ANSWER: -1.46 x 10^-6
Correct! Assuming that you punched the numbers into your calculator correctly.
 
  • #3
Thank you so much!

I believe that I have figured out number one.
I found the energy of each wavelength to be:
4.97 x 10^-19 J
3.62 x 10^-19 J
3.49 x 10^-19 J
and 2.93 x 10^-19 J

Then I found the energy of the electron (2.00 eV multiplyed by 1.6x10^-19, which equals 3.2x10^-19).

Then I subtracted the energy of the wavelength which came closest to 3.2x10^-19 without going over, because I assumed maximum energy is absorbed.

So I did: (3.2x10^-19)-(2.93x10-19) giving me a final answer of: 2.75 x 10^-20 J

Does that last part seem correct?
Also, is my explanation of why it verifies Bohr's model correct?

and..

Hootenanny said:
What does Compton Scattering tell you? (look at the equation).

All I have in my notes is a small section of Rutherford scattering expt, which just says that it was discovered alpha particles bounce off the nucleus, and then my notes have a small bit about Maxwell's suggestion.

I'm trying to search google about Compton scattering, but I'm not really understanding it as we haven't learned any of those equations.

I'm in Physics 30, by the way.

Any more hints would be great! :) Thanks.
 
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  • #4
alexlee33 said:
Thank you so much!

I believe that I have figured out number one.
I found the energy of each wavelength to be:
4.97 x 10^-19 J
3.62 x 10^-19 J
3.49 x 10^-19 J
and 2.93 x 10^-19 J

Then I found the energy of the electron (2.00 eV multiplyed by 1.6x10^-19, which equals 3.2x10^-19).

Then I subtracted the energy of the wavelength which came closest to 3.2x10^-19 without going over, because I assumed maximum energy is absorbed.

So I did: (3.2x10^-19)-(2.93x10-19) giving me a final answer of: 2.75 x 10^-20 J
Looks good to me :approve:
alexlee33 said:
Does that last part seem correct?
Also, is my explanation of why it verifies Bohr's model correct?
Sounds okay to me, but it may be improved by explicitly mentioning that the discrete wavelengths corresponding to discrete energy absorption.
alexlee33 said:
All I have in my notes is a small section of Rutherford scattering expt, which just says that it was discovered alpha particles bounce off the nucleus, and then my notes have a small bit about Maxwell's suggestion.

I'm trying to search google about Compton scattering, but I'm not really understanding it as we haven't learned any of those equations.

I'm in Physics 30, by the way.

Any more hints would be great! :) Thanks.
Physics 30, from what I can gather this would be similar to UK A-Levels, which would make you roughly 17? If that is the case then I wouldn't worry about Compton Scattering. To answer the final question, compare the energies of the photon to and energy of the nucleus (assume that the nucleus is at rest) and think about what would happen classically. Then do the same for the electron.
 
  • #5
Hootenanny said:
Sounds okay to me, but it may be improved by explicitly mentioning that the discrete wavelengths corresponding to discrete energy absorption.

Physics 30, from what I can gather this would be similar to UK A-Levels, which would make you roughly 17? If that is the case then I wouldn't worry about Compton Scattering. To answer the final question, compare the energies of the photon to and energy of the nucleus (assume that the nucleus is at rest) and think about what would happen classically. Then do the same for the electron.

Awesome, thanks! Yup, I'm 17 exactly.
I'll try comparing the energies and see what I can do.
 
  • #6
The only part I could find energy for was the photon, by using E = hc/(wavelength)
The nucleus has no kinetic energy, how would I find it's potential? Or does it have none?
I am not sure how to find the energy of the electron either, since I only know the charge to be 1.6 x 10^-19. Is that equivalent to its energy?

Here are the answers I have come up with:

When the light wave collides with the nucleus:
- wavelength stays the same
- it will bounce off depending on what angle it hits at

When the light wave collides with the electron:
- wavelength increases (energy, therefore frequency, is lost because energy was transferred to the electron)
- wave travels in same direction


I had a 97% final mark in Physics 20; and a 93% so far in Physics 30 -- but this last unit is killing me! It's not making much sense at all... very frustrating. My unit exam is tomorrow - wish me luck, because I'll need it!
 
  • #7
alexlee33 said:
The only part I could find energy for was the photon, by using E = hc/(wavelength)
The nucleus has no kinetic energy, how would I find it's potential? Or does it have none?
I am not sure how to find the energy of the electron either, since I only know the charge to be 1.6 x 10^-19. Is that equivalent to its energy?
Using Einstein's energy-mass relationship the total energy of, say a carbon-12 nucleus, at rest is;

[tex]E_n \approx 12\cdot(931 MeV) \approx 11 GeV[/tex]

Now, if we calculate the associated energy of 570nm a photon;

[tex]E_p = \frac{hc}{\lambda} = \frac{1240}{570} = 2eV[/tex]

So we see that the energy of the nucleus is much greater (~1010) than the energy of the photon. So we have a classical analogy of a ping-pong ball (photon) colliding with a stationary bowling ball (nucleus). And as you correctly say;
alexlee33 said:
When the light wave collides with the nucleus:
- wavelength stays the same
- it will bounce off depending on what angle it hits at
alexlee33 said:
When the light wave collides with the electron:
- wavelength increases (energy, therefore frequency, is lost because energy was transferred to the electron)
- wave travels in same direction[/B]
This is almost correct, but the photon won't generally continue in the same direction. The rest mass of the electron is around 0.511MeV, which is much closer to the energy of a photon.
alexlee33 said:
I had a 97% final mark in Physics 20; and a 93% so far in Physics 30 -- but this last unit is killing me! It's not making much sense at all... very frustrating. My unit exam is tomorrow - wish me luck, because I'll need it!
Good luck! Make sure you come back and let us know how you get on!
 
  • #8
I had my exam this morning, and it went quite well! I studied for 5 hours last night, and thanks to your help too, I find the unit was easier.
My diploma (final) exam is coming up in a few weeks after Christmas break; so I may pop in if I have any more questions from previous units as I study!

Thanks again :)
 
  • #9
Compton Scattering of X-Rays

Hi all. I'm new here. Currently I'm enrolled into first year of physics science undergraduate in a local university in M'sia. I'm currently having some difficulties in my lab report concerning the above title. I've come up with the graph and i was confused. Anyone can help me with this dilemma here please...

The task requires me to determine the change of wavelength using an x-ray unit. The graph i plotted seems very weird. How should be the graph look like. Too bad its a new practical procedure in my course and no seniors had done before.
 

What is the Frank-Hertz experiment and what does it show?

The Frank-Hertz experiment is an experiment conducted to study the energy levels of atoms. It involves passing electrons through a tube of gas and measuring the energy of the electrons before and after they pass through the gas. The experiment shows that atoms have discrete energy levels and that electrons can only absorb or emit energy in discrete amounts.

What is the Compton effect and how does it support the wave-particle duality of light?

The Compton effect is the phenomenon where X-rays or gamma rays scatter off of electrons, resulting in a change in their wavelength. This supports the wave-particle duality of light because it shows that light can behave like a wave (as it can be scattered) and like a particle (as it transfers energy to the electrons).

What is the De Broglie hypothesis and how is it related to the wave-particle duality?

The De Broglie hypothesis states that all particles, including matter, have a wave-like nature. This is related to the wave-particle duality because it suggests that particles can exhibit both wave-like and particle-like behavior, depending on the experimental setup.

How do the Frank-Hertz, Compton, and De Broglie experiments contribute to our understanding of quantum mechanics?

The Frank-Hertz, Compton, and De Broglie experiments all provide evidence for the wave-particle duality of matter and light. This is a fundamental concept in quantum mechanics, which helps us understand the behavior of particles at the atomic and subatomic level.

What are some real-world applications of the concepts explored in the Frank-Hertz, Compton, and De Broglie experiments?

The concepts explored in these experiments have many real-world applications, including in medical imaging (using the Compton effect to generate X-ray images), in the development of electronic devices (using the Frank-Hertz experiment to understand the behavior of electrons in materials), and in the design of quantum technologies (using the De Broglie hypothesis to understand and manipulate the wave-like nature of particles).

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