Sapling HW #6

[Linette Choi 3L]

Hello could someone help explain to me how to solve part B of this problem? I correctly answered A but I am not sure how to go about solving part B.

A.) When a metal was exposed to photons at a frequency of 1.15×1015 s−1, electrons were emitted with a maximum kinetic energy of 3.30×10−19 J.
Calculate the work function, Φ, of this metal.

answer: 4.32×10^−19 J/photon

B.) What is the maximum number of electrons that could be ejected from this metal by a burst of photons (at some other frequency) with a total energy of 6.07×10−7 J?

[Selena Quispe 2I]

Hi! Because you are trying to find the maximum number of electrons that could be ejected you would divide 6.07*10^-7 by the answer you got in part a (4.32*10^-19). I hope this helps!!

[Stephen Min 1I]

Each incoming photon interacts with an electron, so finding the max number of photons will allow us to find the number of electrons. Given 6.07×10−7 J as the total energy, you would divide by the work function of 4.32×10^−19 J/photon. Joules cancel out to give us the max number of electrons.

[Hannah Rim 2D]

Hi! For this problem, you need to divide the amount the total energy, 6.07×10−7 J by the answer you got in the previous question, 4.32×10^−19 J/photon. It may be clearer if you think of this through dimensional analysis. Since your unit for the total energy is in joules, if you multiply 1 photon/4.32×10^−19 J, you will be left with only photons for your answer. Hope this helps a bit!

[Ven Chavez 2K]

The maximum number of electrons ejected would result in 0 kinetic energy because there is no leftover energy. Since you have the work function from the first part of the problem, you divide the total energy given in the question by the work function you got from the first answer. This would produce the max number of electrons ejected. 6.07×10−7 J/4.32×10^−19 J/photon = final answer.