What is the frequency of light that corresponds to the transit...

Question

What is the frequency of light that corresponds to the transition of the electron from the n = 3 to the n = one state of the hydrogen atom?

Accepted Answer

hey everyone in this example, we're given the below electronic transitions for hydrogen atom and we need to find the frequency of light that is emitted. So we need to go ahead and recognize that the term admitted here is describing an electron emission, which we should recall is describing the electron movement from a higher energy level down to a lower energy level. So we would recall that this end term here means energy level or shell number. We also want to recall our symbol for frequency, which we should recognize is represented in units of hertz or inverse seconds, which are equivalent to one another. And we can also interpret our frequency as finding our potential energy of our photon, which we should also recognize is equal to the potential energy times negative one of our hydrogen atom. Our next step is to figure out how to find our potential energy of our photon which is going to be equal to the negative value of our Rydberg constant, multiplied by the difference between Z over our final energy level represented by N. F. And then squared -Z over our initial energy level represented by NI two. We should recall that Z is referring to the atomic number of our element. So in this case it's going to be for hydrogen. And then lastly, we would go ahead and find our frequency by taking our energy from above, which should be a positive value and dividing it by Plank's constant. So let's go ahead and begin part a In part they were transitioning from the initial energy level of three and then our final energy level should be the lower one because it's an emission, which in this case is two for part A. So for part A we're going to find first delta G for a photon which is going to equal negative one times are Rydberg constant, which we should recall is 2.18 times 10 to the negative 18th power jewels. And then this is multiplied by the difference between our atomic number for hydrogen, which on our periodic tables we would see is one over our final energy level squared. So in that case that would be the lower number for an emission, which in part A is going to be too. And this is squared minus and I'll just slide this down so we have room. So this is going to be minus the same thing for Z, which is one for hydrogen or atomic number over our initial energy squared, which foreign emission is going to be the higher number And that is going to be given us three in part a. So what we're going to get is a value equal to negative 3.278 times 10 to the negative 19th power jewels. And so now we can go ahead and because we understand that the energy of our photon is equal to the negative value of our potential energy of our hydrogen atom. We can say that frequency is equal to negative one times our potential energy of a photon negative 3.278 times 10 to the negative 19th power jewels. And we're going to divide that by plank's constant. Which we should recall is 6.626 times 10 to the negative 34th Power jules times seconds. And so because we have two negatives next to each other, this is really going to be positive, which is what we want for our frequency. And we're going to get a result when we take the quotient here Equal to 4.56 times 10 to the 14th power. And our units are going to be jewels canceled out because we have it in the numerator and denominator. So our final unit is seconds. Because second in the denominator, our units are inverse seconds. And so this is again equivalent to hurts. So this would be our first answer for part A as the frequency of the light emitted for our given transition. And if we go ahead and look at Our electromagnetic spectrum, we would see that because we have a power of 10 to the 14th as our frequency. This is going to refer to infrared light that is being emitted by hydrogen atom at this frequency. So moving on to part B, we're going to follow the same steps first finding our potential energy of our photon which again is negative one times are Rydberg constant. We recall is 2.18 times 10 to the negative 18th power jewels. Multiplied by the difference between our atomic number for hydrogen. One in the numerator divided by our final energy level for an emission which is the lower number in part B. That would be four. So that would be four squared minus our atomic number for hydrogen one over our initial energy level, which foreign emission is the higher number given in part B. S six. So that would be six squared. This gives us a value equal to negative 7.5694 times 10 to the negative 20th power jewels. And just so the math is clear here, we're going to be following the parentheses first. So we're going to find that difference first. And for whatever value we get here, we're going to multiply by our negative one times the Rydberg constant to get this as our result. And so the next step is going to be again finding our frequency of light emitted, which again is interpreted as negative one times our potential energy of our photon equal to negative 7.56 94 times 10 to the negative 20th power jewels, which represents as a whole. Our energy of our hydrogen atom. This is divided by plank's constant again which is we recall is 6.626 times 10 to the negative 34th power jewels times seconds. So again we have jewels over jewels, we can cancel jewels out, leaving us with seconds in the denominator. And so what we're going to get here is a value of 1.14 times 10 to the 14th power inverse seconds. And sorry, that should just have a wonder. So this would be our final answer for part B. Here, which again, according to our exponents 10 to the 14 corresponds to infrared light that is being emitted. So moving on to part C in part C first. Again, we want to start out by finding the potential energy of our photon, which again is equal to the negative value of our Rydberg constant 2.18 times 10 to the negative 18th power jewels. Multiplied by the difference between our atomic number for hydrogen. One Over our final energy level foreign emission, which is the lower number given in part C as one. So that would be 1/1 squared -1. Our atomic number for hydrogen over our denominator which we have our initial energy level in an emission which is going to be the higher number in part C given as four. So that would be 1/4 squared So this is going to give us a result equal to negative 2.437 times 10 to the negative 18th power jewels. And so now that we have our potential energy of our photon, we can find frequency by taking in our numerator, the energy level of our hydrogen atom interpreted as negative one times our potential energy of our photon jules. And in our denominator we want to plug in again. The plank's constant, which we recall is 6.626 times 10 to the negative 34th power jewels. Times seconds Again we have jewels over jewels, cancel those out. We're left with seconds as our final answer or our final unit. And this is going to give us a value Equal to 3.08 times to the positive 15th power inverse seconds. Because seconds are in the denominator. So this is our final answer for part C. And as we can tell, looking at our electromagnetic magnetic spectrum for the power of 10 to the 15 hertz. That should correspond to the frequency of visible light which is being emitted by a hydrogen atom for part C. So in part D, we're just going to solve that last. And what we would have first again is our potential energy of our photon equal to the negative value of our Rydberg constant. 2.18 times 10 to the negative 18th powered jewels. Then multiplied by the difference between our atomic number for hydrogen in the numerator, one over our final energy level Given in part D, which is going to be foreign emission. Our lower number given us too. So that would be 1/2. This should be squared minus Our atomic number for hydrogen again, one in the numerator divided by the final energy level. Or sorry, the initial energy level for an emission which is going to be the higher number given as five in part D. So 1/5 squared This is going to give us a value equal to negative 4.578 times 10 to the negative 19th power jewels. And now we can go ahead and find our frequency by interpreting our energy for our hydrogen atom as negative one times our potential energy of our photon From above as negative 4.578 times 10 to the negative 19th power jewels. And then in our denominator we recall plank's constant for the last time as 6.626 times 10 to the negative 34th power jewels times seconds. And so we can go ahead and cancel out jewels over jewels were left with seconds in our denominator. And what we would have is our frequency for part D. Is equal to 6.91 times 10 to the positive 14th power inverse seconds or hurts. And so this is going to be our final answer to complete this example for party. And when we look at our exponent here in Hertz 10 to again as we saw earlier that corresponds on our electromagnetic spectrum to visible light. And so this will complete this example as all of our final answers that our hydrogen atom emitted at each given frequency in parts A through D. So I hope that everything I explained was clear. But if you have any questions, please leave them down below, and I will see everyone in the next practice video.

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What is the frequency of light that corresponds to the transition of the electron from the n = 3 to the n = one state of the hydrogen atom?

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