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OpenStudy (anonymous):
Calculate the wavelength (λ in nm) of the photon emitted when an electron transitions between n = 3 and n = 1 energy levels. I keep getting .00975 but the answer is apparently 102.6nm could someone please explain this to me?
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OpenStudy (somy):
dsinQ=nLambda
OpenStudy (somy):
lambda= ( d * sinQ )/ n
OpenStudy (somy):
ah wait its not what you need, sorry
OpenStudy (somy):
can u show your working?
OpenStudy (jfraser):
0.00975 is the INVERSE of the wavelength. If you divide 1/0.00975 you'll get 102nm
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OpenStudy (anonymous):
Thanks Fraser that's exactly what i needed
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