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Mandarinka [93]
2 years ago
15

The charge on the sphere is monitored as a beam of monochromatic light shines on the sphere. Initially nothing happens. The wave

length of the light is slowly decreased. When the wavelength reaches a certain value, a positive charge is suddenly measured on the sphere. The wavelength is then held constant, and the charge continues to increase at a constant rate. The intensity of the beam is then increased without the wavelength being changed, and the rate of increase of the charge becomes greater.
In a coherent paragraph-length response, describe the cause of the charge on the sphere and the changes in the observations about the charge, in terms of physics principles.
Physics
1 answer:
Solnce55 [7]2 years ago
8 0

Answer:

explanation of this effect  is  the photoelectric effect

Explanation:

Let's describe the process, when light of large wavelength falls, this implies a small energy, according to Planck's equation

           E = h f = \frac{h \ c}{ \lambda}

the energy of the photons is not enough to carry out an electronic transition between two states of the material, when we decrease the wavelength (the energy of the photons increases), the point is reached where the energy of the beam is equal to some energy of a transition, by which the electrons are promoted and since we can see a certain charge, as the atoms are neutral, some electrons must be removed from the material, this is represented in the macroscopic case as the work function of the material, consequently a unbalanced load that is what we can measure.

When we increase the lightning intensity, what we do is that we increase the number of photons and if each photon can remove an electron, by removing the electrons the difference between it and the positive charge (fixed in the nuclei) increases.

We can analyze the interaction of the photon and the electron as a particular collision.

The explanation of this effect was made by Einstein in his explained of the photoelectric effect

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I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

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