Photo electric effect problem.

In summary, the stopping potential for a wavelength of 400 nm is 25.7% of the stopping potential for a wavelength of 300 nm, indicating that the same metal cathode was used in the experiment. To find the work function for the 300 nm wavelength, equations can be set up and solved using the given information.
  • #1
btbam91
91
0
In a photoelectric-effect experiment, the stopping potential at a wavelength of 400 nm
is 25.7% of the stopping potential at a wavelength of 300 nm. Of what metal is the
cathode made?

So I need to find the stopping potential of the metal with a wave length of 400 nm, but first I need to find the stopping potential of the metal with a wave length of 300 nm.


Using:

V(stop)=Kmax/e

Kmax=hf-Eo(work function)

How do I find the work function for the 300 nm?


Thanks.
 
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  • #2
I believe that the problem states that it is the same metal for both cases. That is, the same cathode has 400 nm light, then 300 nm light shone on it, and the result was that the stopping potential was less for the longer wavelength light (which is expected), and in fact was only 25.7% of the stopping potential for the shorter wavelength light.

Sounds like a case of setting up two equations in two unknowns and solving. Write expressions for both stopping potentials, then use the given information about their numerical relationship.
 

Related to Photo electric effect problem.

What is the photoelectric effect problem?

The photoelectric effect problem refers to the observation that when certain metals are exposed to light, they emit electrons. This phenomenon cannot be explained by classical wave theory, leading to the development of quantum mechanics.

What is the significance of the photoelectric effect?

The photoelectric effect provided evidence for the particle nature of light, which was a major breakthrough in understanding the behavior of light. It also led to the development of the concept of photons and paved the way for the development of technologies such as solar cells, photocells, and photomultiplier tubes.

What factors affect the intensity of the emitted electrons in the photoelectric effect?

The intensity of the light, the frequency of the light, and the work function of the metal are the three main factors that affect the intensity of the emitted electrons in the photoelectric effect. Increasing the intensity of the light or the frequency of the light will result in more electrons being emitted, while increasing the work function of the metal will decrease the number of electrons emitted.

How does the photoelectric effect support the wave-particle duality of light?

The photoelectric effect shows that light behaves as both a wave and a particle. The energy of the emitted electrons is directly proportional to the frequency of the light, which is a characteristic of waves. However, the number of electrons emitted is dependent on the intensity of the light, which is a characteristic of particles.

What are the real-world applications of the photoelectric effect?

The photoelectric effect has many practical applications, such as in solar panels, which use the photoelectric effect to convert sunlight into electricity. It is also used in photocells and photomultiplier tubes to detect and amplify light signals. Additionally, the photoelectric effect is used in photography and in the development of digital cameras.

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