Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n =?

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n = 4 to n = 1.

= ? J

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  • 8 years ago
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    We can find the energy (E) from the wavelength (w) by: E = hc/w where

    h is Planck's constant, 6.625 * 10**-34 joule-sec. and c the speed of light

    which is 3.0 * 10**8 meters per sec. so E = hc/w

    We can find w from the lower (L) and upper (U) energy levels (1 and 4)

    by Rydberg's formula: 1/w = R(1/L² - 1/U²) where R is Rydberg's constant

    equal to 10967758 waves per meter for hydrogen. Put these formulas

    together and the energy E = hc/w = hc(1/w) = hcR(1/L² - 1/U²) Combine

    all the constants and we get E = 2.18 * 10**-18 (1 - 1/16) joules

    Then E = (2.18 * 10**-18)(1 - 0.0625) = 2.0437 * 10**-18 joule.

    Whew! Hope this answers your question.

    Source(s): quantum physics in college
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