Question

What frequency of light is needed to eject an electron from a metal whose threshold energy is `"201 kJ/mol"`?

Answer 1

`5.04 xx 10^14` `"Hz"` or `"s"^(-1)`

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This sounds like it's about the photoelectric effect.

The binding energy `phi` is the energy that would have needed to be supplied by an incoming light source in order to remove a desired number of electrons from the surface (in this case, `"1 mol"` of electrons).

It is also defined as the threshold that needs to be overcome in order to eject a specific number of electron(s). Hence, its associated frequency IS the threshold frequency.

If you supply more energy than the required threshold to eject a specific quantity of electrons, the difference in the supplied energy and the binding energy gives the remaining kinetic energy that goes into the ejected electrons.

Therefore, the equation becomes:

`bb(stackrel("leftover kinetic energy")overbrace(K_e) = stackrel("supplied")overbrace(E_"photon"))` `-bb()` `bb(stackrel("required threshold")overbrace(phi))`

`= bb(hnu - hnu_0)`

where `nu_0` is the threshold frequency.

Based on your question, all you need is the threshold frequency `nu_0`, which means:

`phi = hnu_0`

`color(blue)(nu_0) = phi/h`

`=` `[(201 cancel"kJ")/cancel"mol electrons" xx cancel"1 mol electrons"/(6.022xx10^(23)) xx (1000 cancel"J")/cancel"1 kJ"]/[6.626xx10^(-34) cancel("J")cdot"s"]`

`= color(blue)(5.04 xx 10^14` `color(blue)("Hz")` or `"s"^(-1)`