Beryllium has a work function of 5.0 eV. Calculate the minimum frequency of ligh

Question

Beryllium has a work function of 5.0 eV. Calculate the minimum frequency of light that would eject electrons in the photoelectric effect experiment. If light of frequency 8*1015 Hz was used, what would be the maximum kinetic energy of the ejected electrons?

????max = hf - Wo

I know how to calculate the maximum kinetic energy for this question using the formula above but what formula would be used to calculate the minimum frequency needed to eject the electrons?   I am thinking I use the same KEmax equation where 5.0 eV (converted to J) is the work function (Wo) and the KEmax and then solve for f. This is because for photoelectric effect the hf > Wo in order to eject an electron. Does that sound correct?

I get a mimimum KE of 2.41 x 10^15 m is needed if I am doing that correctly.

Explanation / Answer

You can get minimum frequency( needed to eject the electrons) in the case when there is a work function given. You can simply get it by formula :- work function =  h*f(minimum). Else you can't find the minimum frequency as it is an inherent property of atom and can be found only by using atomic structure rules.

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