Cesium metal is frequently used in photoelectric cells because th...

Question

Cesium metal is frequently used in photoelectric cells because the amount of energy necessary to eject electrons from a cesium surface is relatively small—only 206.5 kJ/mol. What wavelength of light in nanometers does this correspond to?

Accepted Answer

Hey everyone in this example, we need to calculate the wavelength of radiation that will provide a photon that would reach the energy threshold of iron at this value given. So we're going to begin by recalling our formula for the energy of a photon, which we would recall is equal to Planck's constant, multiplied by our frequency. We also want to recall our formula where wavelength is equal to our speed of light divided by frequency. So we first need to find frequency. So we're going to reinterpret our formula for energy of a photon and say that that's equal to our energy divided by plank's constant. So finding frequency according to the info and are pumped. We would say that frequency is equal to Our energy threshold given in the prompt is 6.878 times 10 to the negative 19th power jewels. And we're dividing this by plank's constant, which we recall is 6.626 times 10 to the 23rd power or sorry to the negative 34th power jules times seconds. So now we're able to cancel out jewels were left with inverse seconds for frequency which is the proper unit. And this gives us a value equal to 1.03, 8 times 10 to the positive 15th power in units of inverse seconds. So this is step one of our solution. And now we're going to move on to step two to find our wavelength. So we should recall that in our numerator, our speed of light is equal to a value of 2.998 Times 10 to the 8th power m/s. And then in our denominator we're going to recall that Our frequency, which we just calculated above is equal to 1.38 times 10 to the 15th power inverse seconds. So we're able to cancel out inverse seconds with seconds were left with meters as our unit for wavelength. And we get our final value for wavelength equal to 2.888 times 10 to the negative seventh power meters. So this was step two of our solution and this was our final answer for our wavelength of radiation. That gives our photon the energy threshold of 6.87, 8 times 10 to the negative 19th power jewels. So what's boxed in is our final answers? If you have any questions, please leave them down below and I will see everyone in the next practice video.

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