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Hydrogen gas bombarded by electrons

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Homework Statement


What wavelengths will be emitted if you bombard hydrogen gas (in its ground state) with 13 eV electrons.



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The Attempt at a Solution


I talked to my professor about this one and he mumbled something about the ratio of the energies. I know the ground state energy is -13.6 eV and it's getting hit with 13 eV electrons. The ratio is 13/-13.6 = -.956

Or, I was reading that I should add the energies and then find the wavelength. Either way, the wavelength is pretty tiny.....

Can I get a hint at what to do?
 
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Answers and Replies

  • #2
ehild
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The -13.6 eV means the energy with respect to the ionized state, when the electron is not bound to the atom. But hydrogen has many excited levels. You certainly know the formula about the energy of the n-th level.

ehild
 
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  • #3
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I got tiny wavelengths because I didn't convert to joules. lol

Now, you say that -13.6 is the energy in an unbound electron. The formula I have is E = (-13.6 eV)/n^2. If n = 1, E = -13.6 eV. If it's unbound, n = infinity and E = 0. That's the way I understood it, anyways...
 
  • #4
vela
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You two are saying essentially the same thing. Ehild just meant the energy of the ground state is 13.6 eV lower than the energy of the unbound or ionized state of the atom, so if you take the zero of energy as that of the unbound state, then the energy of the ground state is -13.6 eV.
 
  • #5
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You two are saying essentially the same thing. Ehild just meant the energy of the ground state is 13.6 eV lower than the energy of the unbound or ionized state of the atom, so if you take the zero of energy as that of the unbound state, then the energy of the ground state is -13.6 eV.
I see what you're saying. I don't think I was reading the post right. It's 5 a.m., lol. I'm still not sure how to figure out what happens with the gas. My gut says to add 13 and -13.6 because there will be energy left over and figure out a wavelength from that.... If that makes any sense. I really don't understand this stuff...
 
  • #6
ehild
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What is the energy of the electron on the second , third, fourth level (n= 2, 3, 4 ? Is the 13.6 eV enough to excite an electron from the ground level onto the second, third, fourth one?

The electron can jump onto a lower level from an excited one, emitting a photon which energy is equal to the energy difference between the initial and final levels.

ehild
 
  • #7
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What is the energy of the electron on the second , third, fourth level (n= 2, 3, 4 ? Is the 13.6 eV enough to excite an electron from the ground level onto the second, third, fourth one?

The electron can jump onto a lower level from an excited one, emitting a photon which energy is equal to the energy difference between the initial and final levels.

ehild
+13.6 would free the electron. +13 comes close so, um, what happens then? :*)
 
  • #8
ehild
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Remember the Bohr model of the hydrogen atom. You have written in a previous post : "The formula I have is E = (-13.6 eV)/n^2". These are the possible energy values of an electron in a hydrogen atom. The electron can jump to a higher energy level if it gets enough energy from somewhere. A high-energy electron can kick up an electron to a higher energy level. What is the energy difference between the second level and the first level? The third and the first ones? The fourth and the first ones? Is that 13 eV enough to rise the electron to a higher level?

ehild
 
  • #9
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Remember the Bohr model of the hydrogen atom. You have written in a previous post : "The formula I have is E = (-13.6 eV)/n^2". These are the possible energy values of an electron in a hydrogen atom. The electron can jump to a higher energy level if it gets enough energy from somewhere. A high-energy electron can kick up an electron to a higher energy level. What is the energy difference between the second level and the first level? The third and the first ones? The fourth and the first ones? Is that 13 eV enough to rise the electron to a higher level?

ehild
I believe you're telling me to find what energy level 13 eV will bump the hydrogen electrons up to.

[tex]\Delta E = -13.6(\frac{1}{n^{2}_{f}} - \frac{1}{n^{2}_{i}})[/tex]
[tex]\Delta E = 13 = -13.6(\frac{1}{n^{2}_{f}} - 1)[/tex]
This give me a value of 4.761 for n but, since n is quantized, I have to round down to 4. With n = 4, I have Ef = 12.75 eV, so what happens to the other .25 eV?

Am I getting there?
 
  • #10
ehild
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I believe you're telling me to find what energy level 13 eV will bump the hydrogen electrons up to.

[tex]\Delta E = -13.6(\frac{1}{n^{2}_{f}} - \frac{1}{n^{2}_{i}})[/tex]
[tex]\Delta E = 13 = -13.6(\frac{1}{n^{2}_{f}} - 1)[/tex]
This give me a value of 4.761 for n but, since n is quantized, I have to round down to 4. With n = 4, I have Ef = 12.75 eV, so what happens to the other .25 eV?

Am I getting there?
It is much better... Never mind that 0.25 eV. The incoming electron can keep it or it goes to the nucleus. Now, you have the electron on the fourth level. It will not stay there, but jumps down, emitting a photon. It can go to the 3rd or to 2nd or to the first level, then can be transitions from n=3 to n= 2 or n=1. From n=2 to n=1. What are the photon energies?

ehild
 
  • #11
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So, I would use the Rydberg formula to figure out the wavelength emitted by each transition down?


Related question, I did the same problem but instead of the gas being in the ground state, it's in the n=5 state. I get an imaginary number for n. How do I resolve this?
 
  • #12
ehild
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Yes, use the Rydberg formula. It is quite an ugly job to calculate so many wavelengths, I admit.

As for the other problem, was the gas bombarded by electrons of the same energy? If so, this energy is plenty enough to ionize the atom. The Rydberg formula is no more valid when the electron gets free. It can have any (non-negative) kinetic energy.

ehild
 

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