When sodium is excited in a flame, two ultraviolet spectral lines at (wave length symbol)=372.1 nm and (wave length symbol)=376.4 nm respectively are emitted. which wavelength represents photons with

- 372.1 nm -> Higher energy.

- 376.4 nm -> Lower energy.

Explanation:

Hello,

By considering that for a photon the shorter the wavelength the higher the energy, since the two wavelengths are very close, one could distinguish them as follows:

- 372.1 nm -> Higher energy.

- 376.4 nm -> Lower energy.

Furthermore, at lower wavelengths, we find higher frequencies.

- 372.1 nm -> Higher frequency.

- 376.4 nm -> Lower frequency.

Best regards.

b is the correct answer

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where are the choice options?