Question

Question #ac059

Answer 1

`615.9nm`

Explanation:

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).