Questions Regarding the Photoelectric Effect

[RJLiberator]

Homework Statement

We are performing a lab on the photoelectric effect tomorrow and have some pre-lab questions.

1: If an electron must use an energy equal to the work function of the phototube cathode, in order to escape the cathode's surface, how much kinetic energy does it have once it escapes?
2: What is the energy required for an electron of charge -e to travel across a potential difference of -V?
3: IF there are two frequencies of light with v_1> v_2>0, in a light beam shining on a phototube whose work function is x_0 at what level of applied voltage across the phototube will all the electrons which received a photon of frequency v_2 and v_1 be stopped? Which of these voltages are greater?

Homework Equations

c = wavelength * frequency
E=hv, h=plancks constant

The Attempt at a Solution

My answers/guesses:

1: Energy = hv-work function where v is the frequency. This seems simple enough to me, but I have not studied this in quite some time, so I was looking for confirmation here.
2: E=V/d where d is the separation. So this would mean V=E*d and since #1, we see V=(hv-workfunction)*d.
3: Since equation in answer 2 shows us that the higher the frequency the higher the voltage, we conclude that the voltage for v_1 is greater. Not sure how to answer the part on applied voltage...

 

[TSny]

OK for part 1. In part 2, the symbol "E" in the equation E = V/d does not stand for energy. You'll need to recall the correct relationship between potential difference and electric potential energy.

 

[RJLiberator]

I see, I see. That is a good catch on your part.

Let's try this from scratch.

Potential Difference = W/Q where W = work done Q = per unit charge.

We say potentital difference = -V=W/(-e)
So we can say the answer to 2 is V=W/e

Would you agree? Is it that simple?

 

[TSny]

Question 2 asks for the energy required. Did you write an expression for the energy required?

 

[RJLiberator]

Oi, this is not good! You are absolutely correct.

But it shouldn't be too hard from my answer.

eV=W
Work is made up of the kinetic energy, so we could use the answer in problem one to assist us here.
eV = (hv-workfunction)
Or just eV = E

 

[TSny]

Right. The energy needed is (-e)(-V) = eV.

 

[RJLiberator]

That makes sense. Any words of advice on this: "at what level of applied voltage across the phototube will all the electrons which received a photon of frequency v_2 and v_1 be stopped? "

 

[TSny]

Which will have more kinetic energy: electrons kicked out by photons of frequency ν₁ or electrons kicked out by photons of frequency ν₂? You will need a voltage that can stop the most energetic electrons.

 

[RJLiberator]

The higher the frequency, the more energy is needed!
So v_1>v_2, thus v_1 will require greater energy.

 

[TSny]

OK. The wording of question 3 is a little confusing to me. I would work out the voltage V₁ required to stop electrons that are kicked out by photons of frequency ν₁ and also work out the voltage V₂ required to stop electrons that are kicked out by photons of frequency ν₂.

It seems that the first question in part 3 is asking for a single voltage that will stop all of the electrons ejected by either frequency. The second question in part 3 seems to be asking for the two individual voltages.

 

[RJLiberator]

I agree with your assessment.
We just had the lab, it was successful. Your help had solidified my understanding of the (simple) mathematical operations involved here.

 

[TSny]

Great! Hope you enjoyed the lab.