Question #ac059

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Answer 1 · 2016-09-14T19:08:00

$615.9nm$

Energy in atom is transferred as light energy during emission.

Energy of photon of light given by

$E=hf$

Hence the frequency of the emitted photons is

$f=E/h=(3.215xx10^(-19))/(6.6xx10^(-34))$

$=4.871xx10^14 Hz$

Hence the wavelength may be found as :

$c=flambda =>lambda = c/f = (3xx10^8)/(4.871xx10^14)=615.9nm$

(which I think is in the visible blue region of the em spectrum).