Physics Q pack #26 & 29

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Amber9
Posts: 29
Joined: Thu Jul 04, 2019 11:08 am

Physics Q pack #26 & 29

Post by Amber9 » Fri Jul 05, 2019 3:37 pm

How do the explanations of both questions relate to the equation: electron energy (kinetic and potential?) = hf (photon energy) - work function
NS_Tutor_Mathias
Posts: 264
Joined: Sat Mar 30, 2019 8:39 pm

Re: Physics Q pack #26 & 29

Post by NS_Tutor_Mathias » Fri Jul 05, 2019 7:21 pm

#26:
For every photon with an energy greater than or equal to the work function absorbed by the cathode, one electron with a kinetic energy of 0 or greater is ejected. More photons of such energy impacting, the more electrons will be ejected. This question wants you to make the decision between the number of photons absorbed versus the energy of each individual photon. We are not told how much each individual photon has, just that a greater number have an energy above the work function. This energy could be slightly more than the work function or a lot more - we just don't know!

So this question wants you to conceptually understand the question you talked about, and not fall for the trap of thinking that the energies of several photons would combine to eject one electron with a higher energy, or any other possible misunderstanding.

#29
This harps on the exact same idea. Every single photon hitting the cathode will still eject only one electron. Increasing the frequency of each photon would mean increasing the energy of that photon (simply by E=hf), and the resulting kinetic energy (and therefore speed) of the exiting electron would be greater, because the kinetic energy of every ejected electron is simply the energy of the incoming photon minus the work function, and the work function is a constant here.
NS_Tutor_Mathias
Posts: 264
Joined: Sat Mar 30, 2019 8:39 pm

Re: Physics Q pack #26 & 29

Post by NS_Tutor_Mathias » Sat Jul 06, 2019 2:22 pm

Wanted to update this real quick, I made a significant typo ("decreasing frequency") in the original.
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