How Does the Work Function of Cesium Impact Photoelectric Effect Calculations?

[MarkusNaslund19]

Homework Statement

http://www.astro.uvic.ca/~jwillis/teaching/phys215/phys215_assign_2.pdf

It's question 2.

Homework Equations

Work = U/t
E = hf
f = c/lambda
I=Q/t

The Attempt at a Solution

f = c/lambda = (3.00x10^8 m/s)/(530x10^-9 m) = 5.66x10^14 Hz

E = hf = (6.626x10^-34 Js)(5.66x10^14Hz) = 3.75x10^-19 J/photon

Work = 2.00x10^-3 J/s ----> in one second U = 2.00x10^-3 J

#photons = U / E = 5.33 x 10^15 photons

#photo-electrons produced = (5.33 x 10^15)/(10^5) = 5.33x10^10 photo-electrons

Q = (5.33x10^10)(1.60x10^-19 C) = 8.54x10^-9 C

I = 8.54x10^-9 C/s

The thing is I did not use the work function of cesium to solve this. Hence I am unsure if this is right... Any help would be appreciated.

THANKS!

 

[mjsd]

btw, when you say "work", I think you mean "Watts" or more precisely "power".
your method looks good to me.
work function only comes in when you want to know the energy of the photo-electron..but you are only concerned with the amount of them (ie. current) the most you could do is to check that the incoming photon energy can indeed overcome the work function: $$E_\text{photon} > \phi$$. I don't think there is any problem with that

 

[m2003]

I would like to commend you on your attempt at solving this problem. It is clear that you have a good understanding of the relevant equations and have applied them correctly.

However, as you mentioned, you did not use the work function of cesium in your calculation, which could potentially affect the accuracy of your results. The work function is an important factor in the photoelectric effect, as it represents the minimum amount of energy required to remove an electron from a metal surface.

I would suggest revisiting the problem and incorporating the work function of cesium into your calculations. This will give you a more accurate value for the number of photo-electrons produced and the resulting current.

Additionally, it would be helpful to provide units for your final answer to make it easier to interpret. Overall, your approach and understanding of the photoelectric effect are commendable, but it is important to always double-check your work and consider all relevant factors in a problem. Keep up the good work!