What is the work function of a metal?

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SUMMARY

The work function of a metal is a critical parameter in understanding photoelectric effects, with the discussion centering on a calculated value of 5.1 eV. Participants emphasized the importance of considering the signs of values, particularly the negative charge of electrons, when performing calculations. The correct approach involves using the equation |hf| = |V| + |W|, where hf represents photon energy, V is applied voltage, and W is the work function. The energy of the photon in this context is noted as 3.54 eV, indicating that the work function must be less than this value.

PREREQUISITES
  • Understanding of photon energy and its relationship to work function
  • Familiarity with electron-volts (eV) as a unit of energy
  • Basic knowledge of photoelectric effect principles
  • Ability to manipulate equations involving voltage and work function
NEXT STEPS
  • Learn how to calculate work function using the equation |hf| = |V| + |W|.
  • Study the concept of electron-volts and their application in physics.
  • Explore the principles of the photoelectric effect in detail.
  • Practice problems involving photon energy and work function calculations.
USEFUL FOR

Students studying physics, particularly those focusing on quantum mechanics and the photoelectric effect, as well as educators seeking to clarify concepts related to work function and energy calculations.

Bolter
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Homework Statement
Calculate the metal's work function
Relevant Equations
See below
Question:

Screenshot 2020-03-08 at 14.22.59.png


I have tried this and got work function to e 5.1eV

IMG_4062.jpg


My concern is that for these type of questions, do I need to take into account the signs of some values; such as the negative sign for the charge of an electron? Or could I just take the magnitude for all the values

Any help would be appreciated! Thanks
 
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Bolter said:
Homework Statement:: Calculate the metal's work function
Relevant Equations:: See below

Question:

View attachment 258343

I have tried this and got work function to e 5.1eV

View attachment 258344

My concern is that for these type of questions, do I need to take into account the signs of some values; such as the negative sign for the charge of an electron? Or could I just take the magnitude for all the values

Any help would be appreciated! Thanks
You do need to consider signs. But the easiest way to get it right is to use simple logic. You know that the photon energy is having to overcome both the work function and the applied voltage, so you can write |hf|=[V|+|W|.
 
It doesn't look like any of the possible answers is correct. The energy of the photon is 3.54 eV. If there is a photocurrent, the work function of the metal must be less than this.

You could save yourself a bit of time and reduce the chance of an error by spending a few minutes to learn to work with electron-volts instead of joules.
 
haruspex said:
You do need to consider signs. But the easiest way to get it right is to use simple logic. You know that the photon energy is having to overcome both the work function and the applied voltage, so you can write |hf|=[V|+|W|.

Ok noted. I’ll try to remember this the next time I stumble across a problem similar to this.

When I do use this, I get an answer of 1.99ev? Which doesn’t correlate to any of the options given to me :oldconfused:
 
vela said:
It doesn't look like any of the possible answers is correct. The energy of the photon is 3.54 eV. If there is a photocurrent, the work function of the metal must be less than this.

You could save yourself a bit of time and reduce the chance of an error by spending a few minutes to learn to work with electron-volts instead of joules.

Yep have given this a try when working with electron volts and I get a value of work function to be 1.99ev; which doesn’t match up to any of the options like you have mentioned earlier
 

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