What is the toll for an electron to escape a metal surface?

[Jeff97]

Homework Statement

When Uv light of the frequency of 7.5x10^14Hz is shone on the emitter plate, the maximum KE of the emitted electrons is found to be 1.3x10^-19J Calculate the work function.

Relevant Equations

Φ = hf

Φ = hf = 6.626x10^-34x7.5x10^14 = 4.9695x10^-19 J

 

[haruspex]

Homework Statement:: When Uv light of the frequency of 7.5x10^14Hz is shone on the emitter plate, the maximum KE of the emitted electrons is found to be 1.3x10^-19J Calculate the work function.
Relevant Equations:: Φ = hf

Φ = hf = 6.626x10^-34x7.5x10^14 = 4.9695x10^-19 ?

Also had to conver 1.3x10^-19J into eV which = 0.8125eV ?

I don't see why you needed to convert to eV, nor where you have used the KE. Is 4.9695x10^-19 your answer to the question posed?

 

[Cutter Ketch]

Why convert into eV? You found the energy of the photon in J. The maximum energy of the ejected electrons is given in J.

With those two quantities you are almost done. Now, what do you think the work function is?

 

[Jeff97]

I don't see why you needed to convert to eV, nor where you have used the KE. Is 4.9695x10^-19 your answer to the question posed?

I guess. Have I used the wrong equation?

 

[Cutter Ketch]

I guess. Have I used the wrong equation?

Your equation is fine. You just aren’t done. You have calculated the energy of the photon. But you know that is more than the work function because electrons are popping off with significant kinetic energy. So, now, what do you think the work function is?

 

[Jeff97]

Do I take the E(photon)- KE? if so 4.9695x10^-19- 1.3x10^-19J =3.6695x10^-19J

 

[haruspex]

Do I take the E(photon)- KE? if so 4.9695x10^-19- 1.3x10^-19J =3.6695x10^-19J

Yes, except it is wrong to quote so many digits when one of your inputs is only accurate to two sig figs.
The 1.3x10^-19 could be anything from 1.25x10^-19 to 1.35x10^-19. Round your answer accordingly.

It bothers me that you are uncertain about the basis of your calculation. The incoling photon has a certain energy. Some goes into breaking an electron free of the metal plate (that's the work function), some of what is left over becomes KE, and some may be lost in other ways.
We assume the work function is constant, and that what is lost in other ways can be very small. Consequently, we can suppose that the max KE observed is when those other losses are negligible.

 

[Jeff97]

So to correctly solve for work I need to do.

Φ = hf = 6.626x10^-34x7.5x10^14 = 4.9695x10^-19 J = 4.967x10^-19J (rounded)

4.967x10^-19- 1.30x10^-19 =3.667x10^-19J ?

 

[haruspex]

So to correctly solve for work I need to do.

Φ = hf = 6.626x10^-34x7.5x10^14 = 4.9695x10^-19 J = 4.967x10^-19J (rounded)

4.967x10^-19- 1.30x10^-19 =3.667x10^-19J ?

You are still quoting too many digits. In post #1 you quote the max KE as 1.3 10^-19, not 1.30 10^-19. That extra 0 changes things.
This means that after the subtraction you cannot trust the digit in the 10^-21 position. The 1.3 is anything from 1.25 to 1.35. So what is the possible range of results of 4.97-1.3?

 

[ZapperZ]

So to correctly solve for work I need to do.

Φ = hf = 6.626x10^-34x7.5x10^14 = 4.9695x10^-19 J = 4.967x10^-19J (rounded)

4.967x10^-19- 1.30x10^-19 =3.667x10^-19J ?

You got to a toll booth with $20 in your pocket. You left the toll booth with $15 left. How much was the toll?

The incoming energy of a photon is the amount of original money in the pocket.
The "work function" is the "toll" an electron has to "pay" to get out of the metal.
The kinetic energy of the emitted electron is the amount of money left after it has paid the toll.

You need to understand the physics here, not simply plugging-and-chugging blindly without knowing what they are for. You should also get into the habit of (i) solving the problem SYMBOLICALLY in the beginning BEFORE plugging in numbers, and (ii) to be mindful of SIGNIFICANT FIGURES. If this is anywhere near the type of class that I teach, the issue of significant figures is PART of the course, and you may penalized for ignoring it.

Zz.