The work function of cesium metal is 188 kJ/mol, which corresponds to light with a wavelength of 637 nm. Which of the following will cause the smallest number of electrons to be ejected from cesium?
(a) High-amplitude wave with a wavelength of 500 nm
(b) Low-amplitude wave with a wavelength of 500 nm
(c) High-amplitude wave with a wavelength of 650 nm
(d) Low-amplitude wave with a wavelength of 650 nm

Question

The work function of cesium metal is 188 kJ/mol, which corresponds to light with a wavelength of 637 nm. Which of the following will cause the smallest number of electrons to be ejected from cesium?
(a) High-amplitude wave with a wavelength of 500 nm 
(b) Low-amplitude wave with a wavelength of 500 nm 
(c) High-amplitude wave with a wavelength of 650 nm 
(d) Low-amplitude wave with a wavelength of 650 nm

Accepted Answer

Welcome back everyone in this example, we need to calculate the wavelength of radiation needed to eject electrons at a velocity of 18,000 km/s from an iron metal with a work function of 4.5 electron volts. So because this is a problem focused on radiation, we want to recall that we're going to be emitting photons which contain our electrons that are being admitted at the velocity given with the work function, which is the amount of energy that they are exhibiting. And so we want to recall the total energy change for this photo electric effect and that is represented by delta E. And because this is a unit of energy, our unit should be in jewels. And so we're going to recall that to calculate this. We're going to take the energy of our work function, recall that our work function is represented by the symbol thigh which is a symbol here. And it also should have units of jewels since it's a unit of energy. And we're going to add this to the sum of our total kinetic energy of our electron, which again is a unit of energy. So it should be in jewels. Now recall that this is going to be kinetic energy because we're given the velocity of our electron and recall kinetic energy is a energy with regard to the speed of our object. So what we should recognize is that the units here given in the prompt are in kilometers per second and typically velocity should be in units of meters per second. We're given our work function in units of electron volts. And because we need it in jewels, we're going to convert from electron volts to jules. So let's actually begin step one of our solution by converting our work function to the proper units. So we have 4.5 electron volts. We're going to multiply by the conversion factor where we recall that one electron volt has an equivalent of 1.6022 times 10 to the negative 19th power jewels. So we have electrical to line diagonally. We can cancel them out, leaving us with jewels as our final unit for our work function. And this is going to result in a value of 7.2099 times 10 to the negative 19 powered jewels as our work function. And so now that we have that term we can find our kinetic energy of our electron. So that's going to be part two of our solution where we recall the formula. That kinetic energy can be found from taking one half times the mass of our electron multiplied by its velocity squared so M is our massive electron and V is our velocity of the electron. So plugging in what we know, we can say that our kinetic energy of our electron should equal one half times are massive our electron which is something we should recall from our textbooks as the value 9.11 times 10 to the negative 31st powered kg. That's our mass here. And then our velocity We want to recall is given in the prompt as 1800 km/s, which we stated we need to convert from km/s, two m. So we're going to place kilometers in the denominator and recall. The conversion factor that are prefixed kilo tells us that we have 10 to the third power of our base unit meters, allowing us to get rid of kilometers since it's aligned diagonally, leaving us with meters per second. And then we have from our formula the square power here. So first taking the product of our first two terms, we're going to say that our kinetic energy is equal to the product, which should result in 4.55 times 10 to the negative 31st power kilograms for our first two units. And then for our purple value here for our velocity squared, we're going to get a result of three and sorry, that's a three there. So 3.24 times 10 to the 12th power with units of meters squared divided by seconds squared. Since we had to square everything. And so moving forward, we just want to take the product between these two terms so that we can say that our kinetic energy of our electron is equal to a value of 1.4742 times 10 to the negative 18th power where we have units of kilograms times meters squared divided by r squared second units. And we want to recall that when we have units of kilograms times meters squared divided by seconds squared that this is equivalent to one Juul. And so we can really say that our kinetic energy of our electron is actually equal to just 1.4742 times 10 to the negative 18th power jewels. And so now we have both our work function and our kinetic energy of our electron in jewels. So now we're going to find the total change of our photo total energy change of our photo electric effect of these electrons. And so taking our work function which we calculated above as 7.2099 times 10 to the negative 19 power jewels. And adding that to our value for kinetic energy of our electron. We have 1.4742 times 10 to the negative 18th power jewels. And taking the sum here, we're going to get that our energy change in our photo electric effect when our radiation is emitted is going to be 2.19519 times 10 to the negative 18th power jewels. And now that we have our energy change of our photo electric effect, we are going to go into part four of our solution. So this was part three and part two was us finding the kinetic energy. So now we're at part four of our solution where we're going to use that term delta E. The energy of our photo electric effect. And recall that. That is equivalent to Plank's constant. Multiplied by our velocity. Or sorry, our frequency. So this symbol here is frequency of our radiation. And recall that frequency is expressed in units of hertz or inverse seconds, which are equivalent to one another. Now the prompt asks us for lambda, which is our wavelength. And we want to recall that to get to wavelength. We're going to have to recall that the speed of light is equal to our wavelength. Lambda multiplied by our frequency. So we said, this is wavelength and let's keep the colors consistent. So we have frequency in the color gray. And because we know that that is equal to frequency times wavelength, we can say that our frequency isolating that we would have the speed of light divided by our wavelength. And so therefore with that understood, we can say that our energy change of our photo electric effect is equal to Planck's constant. Multiplied by the speed of light which is divided by our wavelength lambda. And so isolating for wavelength to get to our final answer for the pumps, we can say that our wavelength is equal to Planck's constant, multiplied by the speed of light divided by our energy change of our photo electric effect when our radiation is going through. And so we're going to say that our wavelength is equal to Planck's constant. Which we should recall is the value 6.626 times 10 to the negative 34th power with units of jewels times seconds. We're going to multiply by our speed of light which we should recall is 3. times 10 to the eighth Power units of meters per second. I'm sorry, that should just be 110 there. And then we're going to divide by our energy change of our photo electric effect in the denominator Which above we determined is going to be this value here. So we're going to plug that in as 2. times 10 to the negative 18th power jewels. And so now we're going to just cancel our units. So we're going to get rid of jewels with jewels in the denominator and seconds with in for seconds here, leaving us with meters, which is what we want as our final unit of wavelength. And this is going to result in our final answer for the wavelength of our radiation. That is needed to be 9.55 times 10 to the eighth power meters. And because we recognize in our calculation that our smallest amount of sig figs is just one decimal place here, we're going to round our final answer to one sig fig or one decimal place. And so we would have an answer of 9.1 times 10 to the negative eighth power. And sorry, this should be a negative eight with units of meters. So this is going to be our final answer to complete this example, which above will correspond to choice A And again, this is our wavelength of our radiation that is needed to eject our electrons at the velocity of 18,000 km/s with the work function of 4.5 electron volts. So again, what's highlighted in yellow is our final answer. I hope everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

Verified Solution

Key Concepts

Photoelectric Effect

Work Function

Amplitude of Light Waves