The photoelectric work function of potassium is 2.3 eV. If light that has a wavelength of 190 nm falls on potassium, find (a) the stopping potential in volts

Video transcript

Hi everyone. In this practice problem, we're being asked to calculate the stopping potential. We will have a photoelectric experiment with a light of wavelength 230 nanometer falls on sodium. We're being asked to calculate the stopping potential in faults. If sodium has a photoelectric work function of 2.7 E V. The options given are a 1.69 fold, B 2.96 fold, C 2.69 fold and D 1. fold. So we wanna recall that the maximum kinetic energy for photoelectron or K max can be calculated by multiplying half multiplied by M multiplied by V M X squared. This will equals to E multiplied by V knot or E V knot where V knott is the stopping potential in faults and E V knot is the stopping potential in electron fault. The K max can also be calculated by multiplying H F minus five where H is the plant's constant, which is going to be 4.136 times 10 to the power of negative 15 E V seconds. F is just going to be the frequency which will uh which can be calculated by dividing the speed of light C with the wavelength lambda. So C is the speed of light which is going to be 3.0 times 10 to the power of eight m per second. Five is just going to be the photoelectric work function. So M here is the mass V mass is the maximum velocity H is the planks constant F is the frequency and five is the work function C is the speed of light. And lambda is the wavelength of light. And we can actually combine this equation so that we get. So we will be able to calculate K max. So K max will then equals to each multiplied by F and F can be substituted with C divided by lambda minus five. And we can actually substitute all of our known information into this formula right here in order to calculate K M X. So K M X will then be equal to H which is 4.136 times 10 to the power of negative 15 E V seconds. C is going to be 3.0 times 10 to the power of eight m per second. And then divide all of that by Lambda, which is going to actually be given in the problem statement which is nanometer. So you wanna convert nanometer into meter by multiplying 230 with 10 to the power of negative nine m. And minus that with the work function which is 2.7 E V also given in the problem statement. So then that will give us a K max final value of 2.69 E V, which means that E V knot or the stopping potential in electron folds is also going to equal to 2.69 E V. So finally, we can determine that the stopping potential in faults is going to equal to 2.69 fold. So the final answer to this practice problem with the stopping potential in faults is going to be 2.69 fold, which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out at lesson videos on similar topics available on our website and that'll be it for this one. Thank you.