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What I know

Photoelectric effect

When the surface of a metal is irradiated with an electromagnetic radiation of frequency more than the threshold frequency of the metal, then the photon of the electromagnetic radiation gets absorbed by the atom and an electron is ejected.

Absorption spectrum

We can obtain the absorption spectrum for an element by passing a white light through it and the black lines on the obtained spectrum indicate the colours which got absorbed by the element to exite it's electrons from the ground state to a higher energy state

Question

According to what I understand, all the colours on the absorption spectrum having frequency greater than the threshold frequency should become slightly dimmer. This is because the photons having frequency more than threshold frequency will also be absorbed by the element due to the photoelectric effect.

But APPARENTLY this does not seem to be the case, why? Also is my understanding of the photoelectric effect and the absorption spectrum correct?

For example, the work function of potassium is 2.3 eV, and photons with frequency more than that of green colour contain energy more than 2.3 eV. Hence the absorption spectrum for colours like cyan, blue and purple should become dimmer than the rest of the spectrum but it doesn't appear to happen.

The numbers are from the wikipedia article on Visible spectrum

Edit: I realised that for the absorption spectrum, photon may be absorbed to jump from any lower energy state to a higher energy state, not just from ground to higher energy state

Edit 2: Added an example of what I mean

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1 Answer 1

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Absorption lines are due to electronic transitions from a bound state to another bound state. In the photoelectric effect, the electron transitions from a bound state to a free state. This makes absorption edges, where the opacity suddenly jumps up when the photon energy becomes sufficient to free an electron from a particular bound state.

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