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An electron in a hydrogen atom makes a transition from the n = 7 to the...

An electron in a hydrogen atom makes a transition from the n = 7 to the n = 2 energy state. Determine the wavelength of the emitted photon (in nm). Enter an integer.

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Answer #1

For hydrogen atom electronic transistions, 1/ = Rh (1/n12 - 1/n22), where is wavelength of emitted photon, Rh is rydberg constant(=10967700 m-1), n1 is final principal quantum number, n2 is initial principal quantum number.

So,1/ = 10967700 (1/22 - 1/72) = 2518094.388 m-1

So,=1/2518094.388 = 3.97*10-7 m = 397 nm.

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