My PBL Project

Introduction: My question is: Why does a negatively charged metal plate lose charge if a UV light is shined on it? My Project will focus on the Photoelectric effect, in which light strikes a metal surface and it gives off electrons. It will also focus on how the frequency affected how electrons get off and the kinetic energy that is needed to set off energy. Below are the procedures, materials, and scientific principles used for the experiment.

Materials:

I need: Electroscope or other electrometer, Zinc plate, Plastic rod, Piece of wool fabric, UV lamp, Incandescent lamp, Patch cords with 4-mm banana plug (2), and Alligator chips with 4-mm banana plug (4).

Scientific Principle:

The Experiment talks about the photoelectric effect, in which electrically charged electrons will be loose from its material once the charged photons of the light interact with it. What classical physics can account for the behavior of light doesn't apply to this phenomenon. For example, It was said that the intensity of the light and the rate at which the electrons are separated was part of the Photoelectric effect. However, the effect works almost instantaneously once light was shone on a charged metal plate, and that no matter how much intensity the light has, the effect does not work if the light frequency falls below a certain frequency, thus creating the threshold frequency. Planck's equation (E= hf) answered that the energy of a photon can be transferred to a electron in the metal. Also, hft is equal to the work function of an equation. Thus, (KEmax=hf - hft) is the big equation.

(KEmax=hf - hft) =

Maximum kinetic energy= (Planck's constant Ă— frequency of incoming photon) - work function.

eV is used to measure the maximum kinetic energy, and its significance is 1.60 x 10-19 J. Planck's constant equals 6.63 x 10-34 and its symbol is h.

Safety Regulations:

  1. Avoid Overcharging the vane on an electroscope or other electrometer. These instruments are very sensitive and can be easily damaged.

  2. Be certain that the UV is turned off before it is put away.

Electroscope UV flashlight and incadescent

light blub

Plastic rod, piece of wool fabric, 2 patch cords with 4-mm banana

and a zinc plate plugs, and 4 Alligator chips with 4-

mm banana plug

Procedures:

Part 1:

    1. Set the Electroscope in place and connect the zinc (metal) plate to it. Make sure the Electroscope is insulated from the surrounding objects.

    2. Rub the plastic rod with the piece of wool fabric to generate a negative charge on it, and touch the Electroscope on the top with it. The Electroscope should indicate a reading on it to show charge.

    3. Turn on the UV light by connecting the banana plugs to the outlet and shine the UV light over the zinc plate, discharging the plate. Observe what has happened to the readings.

    4. Then, charge the zinc plate again, and this time use the incandescent lamp to to shine on the plate, and observe the difference it had.

Part 2:

  1. Use the same setup as the one on Part 1, only that when charging the plate, direct the rod close (not touching) to the side of the plate and touch the opposite side with your finger.

  2. Hold the UV light close to the plate and observe the readings.

  3. Hold the Incandescent lamp close to the plate and observe it.

Investigation Part

If UV light with low intensity is known to eject electrons from the surface of zinc, how would the rate at which the electrons are ejected change if you used a UV lamp with higher intensity? Explain your answer.

A: The rate at which the electrons are ejected will NOT change when the intensity of the UV lamp changes. According to experimental evidence, if the frequency of the light falls below the threshold frequency of the metal, the no electrons are emitted; no matter how much intensity the light held. Also, if the light frequency is above a certain intensity, the electrons from the metal surface can emit almost instantaneously. Because the maximum kinetic energy of the electrons is independent of the light intensity, but relies on increasing frequency in order to increase.

Is there a maximum- energy photon for which the photoelectric effect no longer occurs on the surface of a metal? For example, can an X- ray with energy equal to 12.4 keV cause an electron to be liberated from the surface of calcium, which has a work function of 2.9? Explain your answer.

A: Yes, because the photon won't be able to reach the required threshold frequency and can not do the photoelectric effect on the metal surface. But according to the example, the maximum energy of the electrons liberated is 1.52 x 10-18 J, which means that photoelectric effect will work on the calcium surface. If the frequency was negative, then the photoelectric effect would not happen at all.

If nickel has a work function of 5.01 eV, what is the minimum frequency of light that can be used to produce the photoelectric effect from a nickel surface?

A: The minimum frequency of light can be found by solving the threshold frequency, which is similar. I calculated it as ft= 8.016 x 10-19/ 6.63 x 10-34. the threshold frequency is equal to approximately 1.209 x 1015, in which the photoelectric effect will work.

If a photon strikes a platinum surface and an electron is liberated with energy equal to 14.0 eV, what was the frequency of the photon if the work function of platinum is 6.35 eV?

A: I added the work function to the maximum energy of the electron, which makes it 3.256 x 10-18. I made it into an equation of 3.256 x 10-18= 6.63 x 10-34 x f, and divided it by Planck's constant. it turned out to be 4.911 x 1015, which is the frequency of the photon.