Photoelectric effect and mercury

In summary, the photoelectric effect will not take place if mercury is illuminated with UV light with a wavelength of 300 nm, since this wavelength is longer than the cutoff wavelength of 250 nm. To calculate the cutoff frequency of an X-ray tube operating at 44kV, you can use the formula frequency = work function/h, but make sure to correctly convert kV to eV before plugging in the values.
  • #1
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Homework Statement



a. Does the photoelectric effect take place if mercury is illuminated with UV light with a wavelength λ = 300 nm? The cutoff wavelength for mercury is 250 nm.


Homework Equations



ft= WF/h(planks constant)
E=h(f)

The Attempt at a Solution



im not sure how to start this because the work function is not given and neither is the frequency i know that i can get the work function if i have the frequency by multiplying it with planks constant, but I am not sure how to get it with just the wavelength.
 
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  • #2
You are given the cut off wavelength, and wavelength and frequency are related.
But you don't need to calculate anything - just a bit of reasoning.

Your light is longer wavelength than the cut off, is this higher or lower energy?
Which would it have to be to eject an electron?
 
  • #3
o ok, so this means that the energy is lower because the frequency would be lower, so photoelectric effect would not take place. is that right?
 
  • #4
Correct, longer wavelength = lower frequency = lower energy
So no emmission
 
  • #5
alright thanks
 
  • #6
How would you calculate the cutoff frequency of an x-ray tube operating at 44kV? I am using frequency = work function/h. I have (44000eV x 1.602E-19J)/1.626E-34 but my answer is wrong. I'm thinking that I converted kV to eV incorrectly, but I'm not sure.
Thanks.
 

Related to Photoelectric effect and mercury

1. What is the photoelectric effect?

The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light. This effect was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905 through his theory of quantum mechanics.

2. How does the photoelectric effect work?

When a photon of light hits a material, it transfers its energy to an electron in the material. If the energy of the photon is high enough, it can overcome the binding energy of the electron and eject it from the material. The ejected electrons are called photoelectrons and the energy required to eject them is known as the work function.

3. What is the significance of mercury in the photoelectric effect?

Mercury is often used in experiments to demonstrate the photoelectric effect because it has a low work function and emits electrons easily when exposed to light. It was also used in early experiments to study the effect and confirm Einstein's theory.

4. How does the intensity of light affect the photoelectric effect in mercury?

The intensity of light does not affect the kinetic energy of the emitted electrons in the photoelectric effect. However, it does affect the number of electrons emitted, as higher intensity light means more photons and thus more electrons can be ejected from the material.

5. What applications does the photoelectric effect have in modern technology?

The photoelectric effect is the basis for many modern technologies such as solar panels, photodiodes, and digital cameras. It is also used in devices like photocells, photomultiplier tubes, and night vision goggles. Additionally, the photoelectric effect is crucial in understanding the behavior of light and electrons, leading to advancements in quantum mechanics and our understanding of the universe.

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