Ionizing Cesium Atom

[sami23]

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?

[chemisttree]

What value of planck's constant did you use? Remember to include units. Keep track of your units!

[sami23]

h = 6.62*10^-34 J.s

[sami23]

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.

[chemisttree]

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