Can Yellow Light Ionize Cesium Atom?

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Discussion Overview

The discussion centers on whether yellow light with a wavelength of 5230Å can ionize a cesium atom, specifically examining the energy calculations involved in the photoelectric effect. The scope includes theoretical calculations and the application of fundamental physics concepts.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant calculates the energy of yellow light using the formula hc/λ and finds it to be approximately 3.797 x 10^-19 J.
  • Another participant states that the energy required to ionize a cesium atom is 6.224 x 10^-19 J.
  • A participant suggests that since the energy required to ionize cesium is greater than the energy of the yellow light photon, it implies that the yellow light does not ionize the cesium atom.
  • There is a suggestion to use Planck's constant in electron volts to simplify calculations, with a specific value provided (4.14 x 10^-15 eV s).

Areas of Agreement / Disagreement

Participants express differing interpretations regarding the energy comparison, with some suggesting that the yellow light can ionize cesium while others argue against it based on the calculations presented. The discussion remains unresolved.

Contextual Notes

Participants have not reached a consensus on the interpretation of the energy calculations, and there are potential ambiguities regarding the units used in the calculations.

sami23
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Cesium is often used in "electric eyes" for self-opening doors in an application of the photoelectric effect. The amount of energy required to ionize (remove an electron from ) a cesium atom is 3.89 electron volts (1 eV = 1.60 x 10-19 J). Show by calculation whether a beam of yellow light with wavelength 5230Å would ionize a cesium atom.

I converted wavelength: 5230*1x10^-10 = 5.23*10^-7m

Then I calculated the energy by using hc/\lambda :
(6.62*10^-34 * 3*10^8)/(5.23*10^-7) = 3.797*10^-19 J

I also calculated 3.89eV*(1.6*10^-19) = 6.224*10^-19 J energy required to ionize Cesium

How can I tell the beam of yellow light ionizes the Cesium atom?
 
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What value of Planck's constant did you use? Remember to include units. Keep track of your units!
 
h = 6.62*10^-34 J.s
 
I think since 6.224*10^-19 J is required to ionize cesium and there is 3.797*10^-19 J of energy in one photon of yellow light, then you can prove it does ionize because 6.224*10^-19 is less than 3.797*10^-19 J, therefore there is enough energy to ionize cesium. Hope that's right.
 
Look at what you just stated...

My earlier comment was directed at using the value of Planck's constant in eV to make the calculation simpler for you...

h = 4.14 X 10-15 eV s
 

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