Calculating Stopping Potential in Photoelectric Effect Experiment

In summary, the conversation discusses a photoelectric experiment where a stopping potential of 2.70 eV is measured with ultraviolet light of wavelength 380 nm. The question asks for the stopping potential in eV when blue light of wavelength 440 nm is used. The formula qV = hc/λ - work function is mentioned, but there is confusion about the units of potential. It is clarified that stopping potential is typically measured in V, but sometimes referred to as electron energy in eV. It is advised to give the answer in V and explain the incorrectness of the question.
  • #1
struggles
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0

Homework Statement


In a photoelectric experiment, a stopping potential of 2.70 eV is measured when ultraviolet light of wavelength 380 nm is incident on the metal. If blue light of wavelength 440 nm is used, what is the new stopping potential in eV?

Homework Equations

The Attempt at a Solution


I think i understand how to do this problem (qV = hc/λ - work function) but am very confused as they have given and ask for the stopping potential in eV. I thought that eV was a unit of energy and that stopping potential (as it is a voltage) is given by the units V. How can the units of a potential be eV?
 
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  • #2
You're right. The stopping potential should really be measured in V, not eV. But since it is electrons you are stopping, people sometimes get sloppy and refer to electron energy (in eV) rather than the potential (in V).
 
  • #3
So i should give the answer for the energy and not the potential?
 
  • #4
The numerical results should be the same. I would five the answer in V, and explain that the question is really incorrect, since potentials are measured in V.
 
  • #5
ok thank you very much!
 

Related to Calculating Stopping Potential in Photoelectric Effect Experiment

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which certain materials emit electrons when exposed to light. This was first observed by Heinrich Hertz in 1887.

2. What are photoelectric effect units?

Photoelectric effect units, also known as photoemissive units, are used to measure the intensity of light by the number of electrons emitted from a photosensitive material.

3. How are photoelectric effect units measured?

Photoelectric effect units are measured in units of amperes per watt (A/W). This represents the number of electrons emitted per second per watt of incident light.

4. What is the significance of photoelectric effect units in science?

Photoelectric effect units are important in many fields of science, including astronomy, chemistry, and physics. They are used to measure and study the properties of light, such as its intensity and wavelength.

5. How do photoelectric effect units differ from other units of light measurement?

Unlike other units of light measurement, such as lumens or lux, photoelectric effect units directly measure the number of electrons emitted from a photosensitive material, rather than the perceived brightness or illuminance of light.

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