The work function of calcium metal is kJ/mol, which corresponds to light with a wavelength of 432 nm. Which of the following will cause the largest number of electrons to be ejected from cesium? (a) High-amplitude wave with a wavelength of 400 nm (b) Low-amplitude wave with a wavelength of 400 nm (c) High-amplitude wave with a wavelength of 450 nm (d) Low-amplitude wave with a wavelength of 450 nm

Video transcript

Welcome back everyone. We're told that a pure metal sample has a work function of 340 kg per mole corresponding to a wavelength of light equal to 320 nanometers. Which of the following will cause the metal sample to lose the most electrons. So we want to recall that our work function is represented by this symbol here and this is describing the minimum energy that is required to remove electrons. So we're told that we have kg per mole as a work function. And what we should also recognize is that for our final answer to figure out which of the following options will cause the metal sample to lose the most electrons. A higher amount of energy spent removing each electron is going to correspond to a greater amount of energy than our work function. And we would recall that this energy should be interpreted in units of joules per photon and this is going to correspond to a high amplitude wavelength which ultimately will cause the greatest number of electrons removed from our sample. And so we can already rule out choices A as well as choice C. Because we see that they correspond to a low amplitude wave and we're only considering wavelengths of high amplitude. So we're between choices B and D. And to begin, we're going to need to convert our work function from units of kilograms per mole. Two units of joules per photon, As we stated because energy is also measured that way. So we're gonna have our 340 kg per mole. Work function. I'm sorry. Let's make that work function simple better. So this is our work function. We want to convert this from killing jewels, two jewels. So we have killed joules in the denominator jewels in the numerator. We're going to recall that our prefix kilo tells us that we have 10 to the third power of our base unit jewels. So canceling out killing jules, we're now going to get rid of our unit molds by introducing the unit photons through avocados number as six point oh 22 times 10 to the 23rd power photons equal to one mole in our numerator. So now canceling out our units of moles were left with joules per photon as our final unit. And this is going to result in a work function in terms of joules per photon equal to 5.64596 times 10 to the -19 power units of joules per photon. So we want to compare this value to what we get for choices B or D. So beginning with choice B. We're going to recall our formula which relates the energy of a photon to plank's constant H. Multiplied by the speed of light. C divided by our symbol for lambda in the denominator for wavelength. And so plugging in what we know, recall that plank's constant H is to value 6.626 times 10 to the negative 34th power with units of jewels times seconds multiplied by the speed of light, which we recall is 3.0 times 10 to the eighth power meters per second, dividing this by our wavelength given for choice be we have a wavelength of nanometers which we want to recall that We need to cancel out our units of meters in the numerator. So we're going to convert from nanometers two m by recognizing that our prefix nano tells us that we have 10 to the negative nine power meters. So now canceling out our units of nanometers, we're left with meters, which we can cancel out with meters in the numerator and seconds as well, leaving us with jewels as our final unit. So this is going to result in an energy per photon To remove each electron equal to 6.626 times 10 to the -19 power jewels per photon for choice b. So this is definitely a higher amount of energy compared to our work function. But let's go ahead and consider choice D. To see if it's even higher than choice B. So what we'll have is that our energy is equal to again, plank's constant 6.626 times 10 to the negative 34th power jewel's times seconds multiplied by our speed of light 3.2 times 10 to the eighth Power meters per second divided by our wavelength given for choice D. As 360 nanometers, which again we want to convert to meters. So recalling again our prefix nano tells us we have 10 to the negative ninth power of our base unit meter, canceling out nanometers were left with meters, which we cancel out in the numerator as well as with seconds leaving us with jewels as our final unit again. And this is going to yield an energy equal to 5.5217 times 10 to the negative 19th. Our jewels per photon. And we can already see that choice D. Is less than choice B. Meaning that for our final answer choice, we would confirm that choice B. 300 nanometers of wavelength of a high amplitude wave would cause the metal sample to lose the most electrons because it resulted in an energy that was higher than our work function in terms of joules per photon as 6.626 times 10 to the negative 19 power jewels per photon of energy spent removing each electron which is much higher than our energy of our work function. So again, B is our final answer. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video