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Propose a mathematical equation for the photoelectric effect using the terms we have discussed in class (work function; incident light; max E of ejected electrons):...

Question

Propose a mathematical equation for the photoelectric effect using the terms we have discussed in class (work function; incident light; max E of ejected electrons): Use the simulation to verify your model and make a graph of the maximum electron energy vs: the light frequency: Selected metal: light wavelength (nm) light frequency electron stopping voltage Ejected Electron KE (eV) (Hz)190250300350450650

Propose a mathematical equation for the photoelectric effect using the terms we have discussed in class (work function; incident light; max E of ejected electrons): Use the simulation to verify your model and make a graph of the maximum electron energy vs: the light frequency: Selected metal: light wavelength (nm) light frequency electron stopping voltage Ejected Electron KE (eV) (Hz) 190 250 300 350 450 650



Answers

The mathematical equation for studying the photoelectric effect is $$h v=W+\frac{1}{2} m_{e} u^{2}$$ where $v$ is the frequency of light shining on the metal, $W$ is the work function, and $m_{e}$ and $u$ are the mass and speed of the ejected electron. In an experiment, a student found that a maximum wavelength of $351 \mathrm{nm}$ is needed to just dislodge electrons from a zinc metal surface. Calculate the speed (in $\mathrm{m} / \mathrm{s}$ ) of an ejected electron when she employed light with a wavelength of $313 \mathrm{nm}$.

In an experiment to student found that a maximum wavelength of 351 nanometers is needed to just dislodge electrons from a zinc metal surface. Let's calculate the velocity in meters per second. Dominic ejected Electron When the student employee light at a wavelength of 313 nanometers. Let's calculate the energy of wavelength of light at 351 nanometers, which is equal to HC over Lambda Lambda is their wavelength of light six point 63 times 10 to the negative. 34 Jules per seconds. Three times 10 to the 8 m per second is their speed of lights and our wavelength is 351. I'm tired of the negative 9 m. This would work out to 5.66 times 10 to the negative 19 jewels. It's calculate the energy of a wavelength of light that is 313 nanometers. See him formula wavelength of light. Here is 313. I'm still the negative 9 m and this will work out too. 6.35 10 of the negative 19 jewels and now we can calculate the excess energy, which would be equal to 6.35 times in the negative, 19 jewels minus 5.66 times 10 to the negative 19 jewels and this would be equal to 6.9 times 10 to the negative 20 jewels. And now let's calculate the velocity oven ejected electron Using our formula here, high energy is equal to one half m. U squared and the energy 6.9 times 10 to the negative. 20. Jules, I have massive an electron 9.109 times 10 to the negative 31 kg you squared. Therefore, our velocity here solving before you is going to be 3.89 times 10 to the fear meters per second.

Yeah, I have friends here it is given there is also evidence for instance to 70 Million microbes that is 2700 and extra. We have to calculate the work function in electron volt and do. And we have to find maximum energy photo for light of the turbulence, 200 million micro that is 2000 Dagestan. What function is defined as as you buy land or not Urgent to see having the value 1-400 electron volt into any strong upon 2700. And this time so it would be four point 59 electoral votes. And in june it is to me 1.6 and 2 10 to the power 90 angels. That is 7.344 10 to the Power -19 2. Energy of photo. You can define as he will end up As she is 1-400 electron volt into an extra upon 2000. Sorry, 20,000. So it is to be 6.2 electron volt and endure. It will be 9.9- 10 to the Power -19 June. That's all. Thanks for watching it.

We're considering a situation where we have the photo electric effect which has an initial maximum kinetic energy ke not of 2.8 electron volts. This is decreased to K one of 1.1 electron volts by increasing the wavelength by a factor of 50%. So the final wavelength Lambda One is equal to the initial wavelength Lambda not plus 1/2 or 50%. The initial wave like landed not this gives us a value of three have slammed, but not so lambda one is three halves lame. Do not. The photo electric effect says that Kay is equal to H times f f being the frequency h being planes constant minus the work function by the frequency, of course, is just the speed of light See, divided by land us. Now we have ah relationship between K Lambda and the work function. While it does ask us to find the work function in part A, I actually think that's easier to do in part B and we can first find the wave like be so since the work function is the same in both cases, right? We can say that Phi if we saw for five is equal to, um h time. See, in this case, Overland did not, minus K not. It is also equal to because it's the same work function in both cases. H time. See Overland a one. I am the one minus K one. Well, we can go ahead and replace Lambda one with its value of three halves. Lambda. Not that we found in the beginning. And Saul for lame. Do not. So what we have here is a church time. See overland, but not minus h times. See over Lambda one, which is three halves, Landon oxidizes. Age time. See, over three halves. Lambda not is equal to que not minus K one. Okay, so further solving for lamb did not we confined? Weaken first pull HC over Lambda. Not out of the expression on the left side to make the syllable easier. So we have HC over lambda not multiplied by one minus 1/3 halves. Well, 1/3 halves is 2/3. So this is one minus 2/3 is equal to que not minus K one. Well, of course, one minus 2/3 is just simply 1/3 so we can do that. simple simplification, and we can also get lame, did not by itself by multiplying it over to the right side of the equation and dividing by K not minus K one. So we end up with is an expression for Lambda. Not that says that lamb did not is equal to 1/3 plank's constant times the speed of light each time see, divided by K not minus K one. Well, since K not in K one or an electron volts. We're gonna use a value of clinks constant that also has electron volts in it. So Plank's constant with respect to an electron volt seconds is 4.1357 times 10 to the minus 15. Plugging all these values in, we find that lambda, not the original wavelength, is equal that 244 times 10 to the minus nine meters or 244 nano meters so we can go ahead and box that it That's our solution for part B. But now we'll see why it was so easy are much easier to do. Part B first, since we want to find the work function Fi, we have this expression here all underlying it in red. We have this expression here that satisfies equal that planes constant tempted speed of light divided by Lambda not minus K Not well, we know everything now in that expression, cause we just found name did not. So, for B, um, five is equal to HC overland, but not minus K not make sure for lambda, not you're using the meters value. So that's 244 times 10 to the minus nine meters. And for, um H you're still using the electron volts. Second version of Plank's Constant 4.1357 times 10 to the minus 15 electron volts. Um, so now I wrote minus five. This is minus K. Not sorry about that. I'm sure you guys caught that typo. So now, plugging these values and we find that the work function fi is equal to 2.3 electron volts weakened box that in this our solution for, um parte a one more little type over there. We did it backwards, but I wrote it as if we did not. This is part a

Affect the magazine planete tendency Game X is equal to a civil Linda minus right whereas his plans constancy is full of light. Lambda is wavelength and s work mentioned so substituting values KMEX is given 2.8 electoral votes and in the June this can be written and H is Planck's constant which is 6.625 multiplied by 10 to the power minus 34 and C is three multiplied by 10 to the power 8 m per second and Lambda minus five. Okay, so from here after solving 4.48 march elevated to the par minus 19 equals two 19 point 875 multiplied by 10 to the power minus 26 divided by lambda minus fight. Okay, no, if the wiggling is increased by 50%. So substituting the new values KMEX is given 1.1 electron also 1.1 milk crab it and to the four minus 1.6 multiplied by 10 to the power minus 19 which will be equals two again at civil and also the 6.625 molecular weight and to the par minus 34 multiplied by 3 to 10 to the Power eight and Linda will become three lambda back to so three lambda who minus five. So from here, after solving 1.76 molecular weight into the par minus 19, it calls to 39.75 multiplied by 10 to the minus 26 divided by three lambda minus five. Okay, so after solving these two values, we get five equals to 2.3 lacto world. So this is the answer for this question. Okay, The work function is obtained as a country at war. So now, moving to the party in which we have to calculate the original wavelength. So no substituting values in the equation. Original one. So we get one point. We have 4.48 multiplayer. We can do the power minus 19. It calls to 19.875 medical related to the par minus 26. They were by Lambda minus 55 is 2.3 electron volt and Bible plane with 1.6 to 10 to the power minus 19. We got this value in the job. So from here, weaken solve for the Lambda and Lambda Will comes out to be portable food, and I know meters. So this is the answer for this question. Okay,


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