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Energy Level

  1. Apr 20, 2009 #1
    Let me know where my logic is off...Much thanks!

    so
    a)
    vyp8k5.jpg
    For T1: Transition -54.4-(-13.6) = 40.8 eV
    For T2: -13.6 - (-6.04) = 7.56 eV
    T3=?
    n4 = n3
    so
    -6.04 + 7.56 = 1.52 eV
    T3 = 1.52 eV
    n4 = 1.52 eV

    b)
    E = pc
    c = 6.63 x 10^-34 JS
    p = E/c
    p = 1.52/ 6.63 x 10^-34 JS
    p=2.29 x 10^33

    c)
    KEmax = hf - wo
    wo = hf , given wo = 4.7 eV
    E = hf , given h = 4.14 x 10^-15 eV

    f = E/h
    f = (1.52)/(4.14 x 10^-15)
    f = 3.67 x 10^14 Hz

    KEmax = (4.14 x 10^-15) (3.67 x 10^14) - 4.7eV
    KEmax = 1.519 - 4.7
    KEmax = -3.18062 ... okay.. doesn't make sense?
     
  2. jcsd
  3. Apr 20, 2009 #2
    Firstly, I'm going to assume this is a model of a Hydrogen atom?

    Right off the bat, looking at your energy levels they appear to be wrong. You're going to need to use the formula En = -13.6eV/n2

    So we can see the energy levels fall into place accoringly:
    n = 1, E = -13.6 eV
    n = 2, E = -3.4 eV
    n = 3, E = -1.5 eV
    n = 4, E = -0.85 eV

    ..and so on.

    Now if you want the transitional energies between to states, you'll need to use the following formula:

    E = E0(1/n22 - 1/n12) = -13.6eV(1/n22 - 1/n12)

    Now after you find the energy of a photon being emitted (or absorbed) by a transition between two states, you want to find the kinetic energy of that photon?

    Kinetic energy of a photon is given by h/[tex]\lambda[/tex], where h is planck's constant. The wavelength of a photon is given by [tex]\lambda[/tex] = hc/E where c is the speed of light, and E is the energy calculated in part a.

    Following these steps will yield the correct results, however make sure your units are consistent. If you are using energies in eV make sure your value for plank's constant is 4.135x10-15 eV*s. If you are using J (in which case your energies would have to be multiplied by 1.602x10-19) then your value for planck's constant should be 6.626x10-34 J*s. Make sure your speed of light is in SI (3x108m/s) and you should be ready to go.

    Hope this helps,

    cheers
     
    Last edited: Apr 20, 2009
  4. Apr 20, 2009 #3

    Redbelly98

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    Or it could be a hydrogen-like ion, where

    En = -13.6 Z2 eV/n2

    Find Z, and you can find the n=4 energy.
     
  5. Apr 20, 2009 #4
    The more I look at it, the more I think it wasn't supposed to be a Hydrogen atom. But I tried this. Does it make sense?

    The original problem:
    1. The diagram above shows the lowest four descrete energy levels of an atom. An electron in the n=4 state makes a transition to the n=2 state, emitting a photon of wavelength 121.9 nm

    (a)Calculate the energy level of the n=4 state.
    (b)Calbulate the momentum of the photon.

    The photon is then incident on a silver surface in a photoelectric experiment, and the surface emits an electron with maximum possible kinetic energy. The work function of the silver is 4.7eV

    (c) Calculate the kinetic energy, in eV, of the emitted electron.
    34y1c1t.jpg



    For part a)
    In order to find the energy level I use the equation E = (hc/lambda), with the given wavelength being 121.9nm and hc = 1.24 x 10^3eV*nm
    So E = [(121.9nm)/1.24 x 10^3eV*nm)]
    E=10.1723

    For part b)
    E=pc, where p is momentum
    Rearranging this I find that p=(E/c)
    where c approximately equals 3 x 10^17
    so p=(10.173/3 x 10^17)
    p=3.39 x 10^-17

    For part c)
    work function of silver is given as 4.7eV
    KEmax=hf-work function
    E= hf=pc
    so replace hf with E
    getting KEmax = E-work function
    KEmax = 10.1723 - 4.7
    KEmax=5.4723
     
  6. Apr 21, 2009 #5

    Redbelly98

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    (b) and (c) look good.

    For (a), you found the correct energy of the photon. But the question asks for the energy of the n=4 level.
     
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