Anonymous
Anonymous asked in Science & MathematicsChemistry · 9 years ago

Atomic Structure Exercises?

What is the energy of a photon given off by one excited hydrogen atom as the electron moves from energy level 3 to energy level 2?

Any help would be greatly appreciated, if you could also please explain how you got the answer that would be amazing! Thank you!

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  • 9 years ago
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    You can do this in two ways.

    A. You can calculate the wavenumber (1/λ) of the emitted photon, using the equation …

    1/λ = R∞ [1/(n2)^2 - 1/(n1)^2]

    … where n1 and n2 are the principal quantum numbers of the energy levels involved in the transition, and R∞ is the Rydberg constant, 1.096 x 10^7 m^-1

    Knowing the wavenumber, you can calculate the associated energy E using …

    E = (hc) x (wavenumber)

    ... where h is Planck’s constant (6.626 x 10^-34 J.s) and c is the speed of light (3.00 x 10^8 m/s).

    OR

    B. You can go direct to the energy lost during the transition, using ...

    E = Ry [1/(n2)^2 - 1/(n1)^2]

    … where Ry is the Rydberg unit of energy, equal to 2.179 x 10^-18 J.

    ----- ----- ----- ----- -----

    Obviously, if you know the value of Ry, the second method is a little shorter.

    E = Ry [1/(n2)^2 - 1/(n1)^2]

    = (2.179 x 10^-18 J) [ (1/2^2) - (1/3^2)]

    = (2.179 x 10^-18 J) [ 1/4 - 1/9]

    = (2.179 x 10^-18 J) [0.2500 - 0.1111]

    = (2.179 x 10^-18 J) [0.1389]

    = 3.027 x 10^-19 J

    The energy carried away by the photon is 3.027 x 10^-19 J.

    ----- ----- ----- ----- -----

    This photon from the n=3 ---> n=2 transition is the H-α photon, with a wavelength of 656 nm in the red region of the visible spectum.

    I hope this is of some use to you.

    see http://en.wikipedia.org/wiki/Balmer_series

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