What would the minimum work function for a metal have to be for v...

Question

What would the minimum work function for a metal have to be for visible light (380–750 nm) to eject photoelectrons?

Accepted Answer

Hi everyone. In this practice problem, we're being asked to determine the minimum value of a work function. We will have an ultraviolet light with a wavelength range of nanometer to 400 nanometer incident on a lithium surface. We're being asked to determine the minimum value of the work function for the lithium surface. In order for it to be able to eject electrons, the options given R A 1.24 E V B 3.10 E V C 4.13 E V and lastly D 3.1 E V. So in order for us to be able to solve this problem, we will have to actually look at the threshold frequency. So threshold frequency or F T H will have to be used in order for us to be able to solve this problem. So we want to recall that the uh formula for actually calculating the work function or will be equals to H multiplied by F. And in this case, you want to employ the F which with FT H which is going to be the threshold frequency five is going to be the work function H is going to be the planks constant. So H is going to just be 4.136 times 10 to the power of negative 15 E V seconds. And after the H is the threshold frequency at the same time F T H will be equals to C divided by lambda T H where C is just the speed of light which is three times to the power of eight m per second. And lambda T H is just the threshold wavelength, we can substitute our F DH with this formula into our work function formula. So that we get uh Phi to be equals to H multiplied by after H with uh can be substituted with C divided by lambda DH. So Phi will be equals to H multiplied by C divided by lambda T H. So from this equation, we can conclude that the work function will actually be minimum when the wavelength is maximum because it is inversely correlated. So the larger the lambda is the smaller the work function is which is what we are interested to find, which is the minimum value of the work function. So lambda T H for the given range of 100 nanometer to nanometer will actually be equals to 400 nanometer, which is what we're going to be using for this formula. So then I can then be calculated by a multiply H which is 4.136 times 10 to the power of negative 15 E V seconds with C which is three times 10 to the power of eight m per second. And then dividing that with the lambda T H which is nanometer. But we have to convert that into meter by multiplying 400 with 10 to the power of negative nine m. And then calculating all of this, we will get our five value in terms of E V which will come out to be 3.10 E V. So 3.10 E V will be our minimum value for the work function for the lidum surface to eject electrons. And that will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem. So that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our videos on similar topics available on our website and that'll be it for this one. Thank you.

What would the minimum work function for a metal have to be for visible light (380–750 nm) to eject photoelectrons?

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