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A photon interacts with a ground state electron in a hydrogen atom and is absorbed. The electron is ejected from the atom and exhibits a de Broglie wavelength of 3.188×10−10 m. Determine the frequency (in hz) of the interacting photon.

  • Chemistry -

    Yes! I foun the correct answer. It's 7.157e15

  • Chemistry -

    can you please help to explain the formula

  • Chemistry -

    I have tried everything but don't know how the formulas go if you could please give a formula i would be very thankful

  • Chemistry -

    Selly could you please give the formula

  • Chemistry -

    First you find the Momentum (P) from the de Broglie eqn: P = h/lambda, where h is the Plancks constant.
    Then you find the velocity (v) from the eqn: P = m*v, where m is the Electron Mass. Then substitute v in the KE eqn: KE = 0.5*m*v^2.
    Add this KE to the Ionization Energy of H from the periodic table to find the Photon Energy.
    Then Photon Energy = h*nu, where nu is the desired frequency ..

    P.S: Don't forget to convert the Ionization Energy from eV to J.

  • Chemistry -

    First find: E = P^2/2Me + E(first ionization)

    P= h / BroglieWavelength

    E(first ionization)=21.7*10^-19

    Me= 9.1*10^-31

    h= 6.626*10^-34

    Once E is found the find frequency:

    frequency(hz)= E / h

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