This
electronic transition would result in the emission of a photon with the highest
energy:
4p
– 2s
<span>This
can be the same with the emission of 4f to 2s which would emit energy in the
visible region. The energy in the visible region would emit more energy than in
the infrared region which makes this emission to have the highest energy.</span>
The answer is a ) electromagnetic waves
The greater the energy, the larger the frequency and the shorter (smaller) the wavelength. Given the relationship between wavelength and frequency — the higher the frequency, the shorter the wavelength — it follows that short wavelengths are more energetic<span> than long wavelengths.</span>