Photoelectric effect question, help , thanks

In summary, the conversation discusses the photoelectric effect and determining the value of h/e from experimental data. The accepted value for h/e is 4.14x10-15, and the percentage difference between the experimental value and accepted value can be calculated using the formula 100·(experiment-accepted)/accepted.
  • #1
RicXeoR
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Photoelectric effect question, help please, thanks!

In the standard experiment the stopping voltage is measured for several wavelengths of incident light. In this apparatus the photocathode is metallic potassium. Suppose that the apparatus has not been used for some time and that its calibration is uncertain. When monochromatic light with a wavelength of 569 nm shines on the potassium surface, the photoelectric current is stopped by a retarding voltage of 0.10 volts. When light with a wavelength of 405 nm is used, the stopping potential is 0.99 volts.
a) Determine from these data the value of h/e.
This i already got: 4.16862E-15 V*s

b) How does the value of h/e determined from these data compare to accepted value? Find the percentage difference between the value value from these data and the accepted value. [ 100·(experiment-accepted)/accepted ]
***I'm stuck here, e.g. what's the accepted value if part a is the experimental value, or is it the other way around, thanks: Can anyone please help me with this part of the problem, thanks yall!
 
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  • #2
Both h (Planck's constant) and e (elementary charge) are constants in nature, so the "accepted value" (maybe you mean expected value?) is just h/e = 6.63x10-34/1.6x10-19 = 4.14x10-15. Compare that to the experimental value using the method that was already given to you.
 
  • #3
yep, you do exacly that.
 

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to light of a certain frequency. This effect was first discovered by Heinrich Hertz in 1887.

2. How does the photoelectric effect work?

The photoelectric effect occurs when a photon of light strikes a material and transfers its energy to an electron, causing the electron to be ejected from the material. The energy of the photon must be equal to or greater than the material's work function for this to happen.

3. What is the significance of the photoelectric effect?

The photoelectric effect is significant because it provided evidence for the particle nature of light and helped to confirm Albert Einstein's theory of the photoelectric effect, for which he was awarded the Nobel Prize in Physics in 1921.

4. What are some real-world applications of the photoelectric effect?

The photoelectric effect is used in many modern technologies, including solar panels, photodiodes, and digital cameras. It is also used in photocells to control lighting and in phototransistors to detect light.

5. How does the photoelectric effect relate to the wave-particle duality of light?

The photoelectric effect demonstrates the dual nature of light, as it can only be explained by considering light as both a wave and a particle. The wave nature of light is responsible for its frequency and the particle nature is responsible for the transfer of energy to electrons in the material.

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