Can Yellow Light Ionize Cesium Atom?

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Cesium atoms require 3.89 electron volts (6.224 x 10^-19 J) to be ionized. A yellow light beam with a wavelength of 5230Å produces energy of approximately 3.797 x 10^-19 J per photon. Since the energy from the yellow light (3.797 x 10^-19 J) is less than the energy needed to ionize cesium (6.224 x 10^-19 J), it cannot ionize the cesium atom. The discussion also highlights the importance of using consistent units, particularly Planck's constant, for accurate calculations. Ultimately, the yellow light beam does not have sufficient energy to ionize cesium.
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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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