The answer is light is emitted as eletronic transition ours from higher energy state to lower energy state.
Energy is 9.16×10^−20J frequeny is 1.38×10^14 s^−1 wavelength is 486nm
The Bohr Model for the energy of an electron in a hydrogen atom gives us:
E=−13.6/n^2 eV
Where n is the principle quantum number.
For the transition
7→4 the difference in energy is given by:
ΔE=−13.6/7^2−[−13.6/4^2]eV
ΔE=(13.6/16)−(13.6/49)eV
ΔE=0.85−0.277=0.572eV
To convert this to Joules we multiply by the electronic charge:
ΔE=0.572×1.6×10^−19=9.16×10^−20J
To find the frequency of the emitted photon we use the Planck expression:
ΔE=hν
∴ν=ΔEh=9.16×10^−20 /(6.626×10^−34) s^−1
ν=1.38×10^14 s^−1
For wavelength c =fλ.
λ=(3*10^8m/s)/(1.38×10^14 s^−1)
=486nm.
Learn more about the eletronic transition here:
https://brainly.com/question/11299441
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