Molybdenum metal must absorb radiation with an energy higher than 7.22 * 10-19 J ('energy threshold') before it can eject an electron from its surface via the photoelectric effect. (c) If molybdenum is irradiated with light of wavelength of 240 nm, what is the maximum possible velocity of the emitted electrons?

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Hey everyone in this example, we need to calculate the maximum possible velocity of the emitted electron. If iron is irradiated with a light with the following wavelength and the energy threshold via the photo electric effect is this given value here. So our first step is to call our formula for the energy of a photon, which we would recall is equal to Planck's constant, multiplied by our speed of light and then divided by our wavelength. And so calculating for our energy, we would get that energy is equal to Planck's constant. We recall is 6.626 times 10 to the negative 34th power in units of joules times seconds. Our speed of light is 3.0 times 10 to the eighth power meters per second. And then in our denominator from the prompt, we're going to plug in our wavelength given as 1. nanometers. And because we want units of meters to cancel out in the numerator, we're going to convert our wavelength two m by recalling that for one nanometer, we have 10 to the negative ninth power meters. So canceling out our units, we can get rid of nanometers meters as well as seconds with inverse seconds leaving us with jewels as our final unit of energy. And what we would get is that our energy of our photon is equal to a value. And actually before we go any further, we need to correct the wavelength that I plugged in. So this would be actually 156 nm. And just so it's clear We have units of jewels here as our final unit. So recalculating for the energy of our photon, we're going to get 1.273 times 10 to the negative 18th power jewels. Now, according to the prompt, were given an energy threshold via the photo electric effect of this value here. So now we want to take the energy difference from the minimum thresholds. And in doing so this is going to give us our kinetic energy. And so what we're going to take is the value from the prompt. So we would have that K. E. Is equal to the difference between Our larger energy value which is 1.273 times 10 to the negative 18th power jewels that we found above Subtracted from the energy value from the prompt of 6.878 times 10 to the negative 19th power jewels. And this difference here gives us our kinetic energy equal to a value of 5.856 times 10 to the negative 19th power jewels. And actually that six should be a two for our kinetic energy. Now, our next step is to recall that our kinetic energy is equal to one half times the mass of our electron multiplied by velocity squared. So we should recall that the mass of an electron Is also equal to a value of 9.109 times 10 to the negative 28th Power grams. Now for our kinetic energy equation we need mass to be in units of kilograms. So we're going to convert from g of our electron two kg. So for one kg we would recall we have 10 to the third power grams were able to cancel out Graham's leaving us with kilograms as our massive our electron. And that's going to give us a value of 9.1 oh nine times to the negative 31st power kilograms. And so now that we have the mass of our electron in the proper units, we're going to go back into calculating for our velocity of our electron. So isolating our equation of kinetic energy for velocity, we would go ahead and say that velocity is equal to the square root of two times our kinetic energy divided by the mass of our electron. And so we're going to calculate for this so that we have velocity equal to in our numerator two times our kinetic energy which above we stated was the difference between our energy value given in the prompt and from our calculated energy of our photon, we said that that above is equal to this value here. So we're plugging this in as our kinetic energy 5.82 times 10 to the negative 19th power jewels. And this will then be divided by our mass of our electron which above we found to be in kilograms as 9.109 times 10 to the negative 31st power kilograms. And this is going to give us our velocity equal to a value of 1.134 times 10 to the six power. And we should recall that velocity is in units of meters per second. And just as a correction, there should be 1.130 times 10 to the six power meters per second. So this would be our final answer for the velocity of our electron that is emitted. I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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Key Concepts

Photoelectric Effect

Energy of a Photon

Kinetic Energy of Emitted Electrons