Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ionizing energy

  1. Dec 3, 2005 #1
    for the following question:
    a proton capatures a negative muon (mass=207me). Find the ionization energy of the atom.

    my problem:
    the answer in my textbook says to use 13.6eV for the E1 ionization energy, but why? i thougth that that value is only true for hydrogen atoms, but that question doesn't seem to have anything to do with a hydrogen atom...
     
  2. jcsd
  3. Dec 3, 2005 #2
    a hydrogen atom is is made of a proton and an electron. Your text book seems to think that a system made of a proton and a negative muon has the same ionization energy that you attribute to a system made up of a proton and an electron. Since your textbook is setting the question, I wouldn't argue with it. If you ever have to find the ionzation energy of such a system for yourself, and can't, then make a post asking where to find it.
     
    Last edited: Dec 4, 2005
  4. Dec 4, 2005 #3
    thank you very much!!! :)
     
  5. Dec 5, 2005 #4
    @@
    i just rethought this question, and there's something i think that's weird~
    the original ionizing energy has to do with mass, and the muon's mass is 207 times bigger, which is a lot~
    so isn't the ionizing energy affected a lot too?
     
  6. Dec 5, 2005 #5
    perhaps you should put the question as it is worded by the textbook, and incude as worded in the textbook any information given. But I agree that the ionization energy would be different although I'm not confident.
     
  7. Dec 5, 2005 #6
    hmm... that is the original question~
     
  8. Dec 5, 2005 #7
    sorry, of course, I was half asleep when I wrote the last post and didn't read carefully. my fault :P. I regret I can't help any further.
     
  9. Dec 6, 2005 #8
    that's alright~ thanks for you original suggestion! :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook