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Topic: Series Limit  (Read 1968 times)

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Offline Coastie17

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Series Limit
« on: September 20, 2012, 08:35:28 PM »
Just need an answer check. Thank you.

Determine the energy of a photon associated with the series limit of the Humphreys (n=6) series.

v-tilde = 109680[(1/62) + (1/∞2)]cm-1 = 3046.7 cm-1 (100 cm/1m) = 304670 m-1

wavelength = 1/v-tilde = 1/304670 m-1 = 3.028 x 10-6 m

c/wavelength = v  :rarrow: (299792458 m/s)/3.28 x 10-6 m = 9.14 x 1013 s-1

E = hv = (6.626 x 10-34 J s)(9.14 x 1013 s-1) = 6.06 x 10-20 J

Offline Borek

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Re: Series Limit
« Reply #1 on: September 21, 2012, 04:21:15 AM »
Logic looks OK. 3.028x10-6 m is wrong, but later you use a correct value of 3.28x10-6 m, so it is probably just a typo.

Not that I checked numerical values.
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