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Quantum physics

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A hydrogen atom has an electron in the fundamental state.
    a. Show that a radiation with λ = 50 nm will ionize the atom.
    b. What will be the excess kinetic energy of the electron in joules?
    Round up your answer to the nearest hundredth.

    2. Relevant equations
    1/lambda= R(1/n^2-1/k^2) (im not sure if this is the equation to be used)

    3. The attempt at a solution

    1/5x10^-8= 1.097x10^7(1/1^2-1/n^2)
    2x10^7=1.097x10^7(1/1^2-1/n^2)
    1.823154057=1-1/n^2
    .8231540565=1/n^2
    1.214839424=n^2
    1.1= n
    how would this show that atom ionizes?
    b. ?
     
  2. jcsd
  3. Feb 15, 2009 #2
    If the value of 'R' is correct ,then 'n=1' because n=always integer ,so after that I think you can able to calculate it what's ans. of part 'b'...
    see, ionized atom means 'neutral atom minus an electron' ,and this will happen when the electron has +ve energy ............

    ........n=1,2,3.....are the possible energy levels .........there are no levels in-between them........so if you have got n=1.1 ........is it possible that the electron is bound ?

    think about it............
     
  4. Feb 16, 2009 #3
    ooo ok. i understand the question now, but would 1.1 just round off to 1? and if it would how is that ionized, if the electron has -eV? and if it doesnt round off then what would happen because i know only integers are for n.
    and for b? because i know to find the energy of the orbit is -13.6 eV/n^2 but its asking for kinetic energy, and im not sure if thats for kinetic energy.
     
  5. Feb 16, 2009 #4
    Total energy=neu*h,where 'neu'=c/lamda...after that if you just subtract 13.6 energy (if fundamental state means n=1) then you will get the kinetic energy of the electron ..... but why? Don't look at the given relation....U have a question ,ans.. it..
    All the statements may be or may not be correct........
     
  6. Feb 16, 2009 #5
    ok.
    so what if its in a bound state,
    doesnt it need to jump from a bound state to a free state to be an integer because it needs to have a positive value.
    and i used the formula that you showed me, but what is h?
    i just did neu=c/lambda-13.6 and i got 1.1 but that is not the kinetic energy cus all "neu" is, is the frequency right so then what is that?
     
    Last edited: Feb 16, 2009
  7. Feb 16, 2009 #6

    Delphi51

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    Homework Helper

    Another way of looking at this is to convert the -13.6 eV of the n=1 state into Joules. The ionization energy is +13.6 eV, enough to bring it up to zero or beyond.
    Also find the energy of the photon in Joules. This is greater than the ionization energy, so part 1 is done. Just subtract to get the "excess energy".
     
  8. Feb 16, 2009 #7
    ok. i got that. but how do i get the energy of the photon?
    is it h(c/lambda)?
    c= speed of light
    lambda= wavelength?
     
  9. Feb 16, 2009 #8

    Delphi51

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    Homework Helper

    Yes E = h(c/lambda) or hf.
    Momentum = h/lambda or hf/c
     
  10. Feb 16, 2009 #9
    ok i think i got it. how does this look
    4.14x10^-15(c/5x10^-8)= 24.84 eV
    24.48-13.6= 11.24 eV of excess kinetic energy
    and it would ionize because it is now positive.
    and for B would i just do the same thing but in joules?
    6.6x10^-34(c/5x10^-8)= 3.96^-18
    3.96^-18-2.178961169x10^-18= 1.78x10^-18 J of excess kinetic energy?
     
  11. Feb 17, 2009 #10

    Redbelly98

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    Looks good. Alternatively, one could convert the 11.24 eV from (A) into Joules.

    By the way, a useful number to keep handy in your notes (or even memorize) is
    h c = 1240 eV nm​
    So for example
    Ephoton = h c / λ
    = (1240 eV nm) / (50 nm)
    = (1240/50) eV ( Cancelled the nm/nm units )
    = 24.8 eV
    It saves on having to carry around a lot of exponential factors like 10-15 and so on.
     
  12. Feb 17, 2009 #11
    o ok that helps a lot and saves a lot of time thanks :)
     
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