The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency 1.45 * 10^15 Hz. Express the answer in electron volts.

Video transcript

Hi everyone. In this practice problem, we're being asked to calculate the maximum kinetic energy. We'll have an infrared radiation of frequency 2.52 times 10 to the power of 15 Hertz causing an electron to eject from a metal surface. We're being asked to calculate the maximum kinetic energy of the electrons ejected in electron faults. If the photoelectric threshold wavelength of the metal surface is 500 nanometer. The options given are a 4.82 EVB six EVC 2.48 EV and lastly D 7.94 EV. So we wanna recall the equation where the speed of light or C can be calculated by multiplying the threshold frequency with the threshold wavelength or lambda. So speed of light or C is just three times 3.00 times 10 to the power of 8 m per second. F the itch is the threshold frequency and lambda th is the lambda or the threshold wavelength. So in this case, we can rearrange this equation so that we get FTH to, to be equal to C divided by a lambda. Th because we are given the information for the trace threshold wavelength and we want to find the threshold frequency. So F DH will then be equal to 3.00 times 10 to the power of 8 m per second. And then the lambda th is given to be 500 nanometer but we want to convert that into meter. So we wanna multiply 500 with 10 to the power of negative nine to get it into meter. So then calculating this, we will get FTH or the threshold frequency to be equal to 6.00 times 10 to the power of 14 Hertz. So next, we want to recall the um formula to calculate maximum kinetic energy where K max can actually be calculated by multiplying half M and PX squared or at the same time KMX can also be calculated by multiplying H which is the plan constant with F which is the frequency minus five. In this case, um at H or uh plan constant is just a constant which is going to be 4.136 times 10 to the power of negative 15 ev seconds. And then F is going to be the frequency which in this case, we are going to look at the threshold frequency. So fo equals to F DH P five is the work function M is the mass and V max is the maximum velocity. So at the threshold frequency F equals to F DH P max will tend towards zero. So I'm gonna write that down at F equals to FTH, then V max will be equals to zero. So in this case, we want to equate this two equation where we have half MP max squared equals to HF minus five. And input this for uh this situation or this condition right here where at F equals F DH three max equals zero. So the left side will then be equals to zero and the right side will equal to H multiplied by F DH minus five. So in this case, we can then get the equation where Phi will equals to H multiplied by F DH. And we can actually calculate what the P five value is. So Phi will then equals to H is a constant which is 4.136 times 10 to the power of negative 15 EV, multiplied by seconds multiplied that with our uh threshold frequency, we just calculated which is 6.00 times 10 to the power of 14 Hertz or one over seconds. And that will give us a fire value or photoelectric work function of 2.48 EV. So next, using this, we can actually then calculate our maximum K max or maximum kinetic energy to be able to solve this problem. So then going back into our formula that we have here, we can then use that formula where KMX will then equals to HF minus five where in this case, H is still 4.136 times 10 to the power of negative 15 ev seconds. F is now going to be the frequency that is uh given in the problem statement which is 2.52 times 10 to the power of 15 Hertz, 2.52 times 10 to the power of 15 Hertz. And we want to minus our five value which is 2.48 EV, which we just calculated this will then give us our K max value after doing our calculations to then be equals to 7.94 EV. So the maximum kinetic energy or KMX is come going to come up to 7.94 EV, which is going to be the maximum kinetic energy of the electrons ejected in electron faults. If the photoelectric trashed hat wavelength of the metal surface, that is in the problem statement is 500 nanometer that will also correspond to option D in our answer choices. So option D will be the answer to this particular practice problem. So that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website and that'll be it for this one. Thank you.