One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (b) Determine the initial and final values of n associated with this emission.

Question

Accepted Answer

Welcome back everyone in this example, we have radiation with a wavelength of 92.315 nanometers emitted by a hydrogen atom were asked to determine the initial and final energy level values that correspond to this emission recall that an electron emission is going from a high energy level to a low energy level. And again, N is our energy level where our electron resides at for our hydrogen atom. Now, next we want to visualize our electromagnetic spectrum since we're dealing with radiation which corresponds to infrared radiation on our spectrum. So I'll paste that below. And we can see based on our spectrum that we have values for energy where higher energy is associated towards the left hand direction of our periodic table because it says energy increases in that direction. And then we have our values for recall frequency represented by this symbol here and then for wavelength here where frequency is measured, recall in hertz. So just that's visible frequency is measured in hertz, which we understand are equivalent to inverse seconds. And wavelength is represented in units of meters or nanometers by the symbol lambda. So because we understand that we're undergoing an electron emission according to the prompt, we're going to likely start out at a final energy level. That is highest in energy which because we're focusing on the portion of our electromagnetic spectrum where infra red radiation is located, we're going to say that the highest energy level can likely be where UV light is located at around 10 to the negative eighth power meters Notice that based on our visible spectrum which is more towards the right of our electromagnetic spectrum. So it will have a lower change in energy ranges at about 10 to the negative six power meters on our electro magnetic spectrum. So it wouldn't be a likely choice for our final energy level of our electron emission. And we can see that infrared radiation is further towards the right on our spectrum. So it also is lower in energy based on our electron emission. So it's going to have a smaller change in energy. So just to write everything out, we're going to say that our change in energy that is highest. It's associated with ultraviolet light because as we see, it's further to the right. So it has the highest energy. And that will mean that that is going to be our final energy level in our electron emission. So it ranks number one out of visible light and infrared radiation. Whereas we have a smaller energy change attributed to visible light since it's lower in energy. So, it would be the second option for our electron emission as the final energy level as well as infrared, which has the smallest change in energy because its furthest to the right. And that would be the third option for our final energy level of our electron emission. So, what we want to do is calculate the change in energy to sulfur what our initial energy level where our electron starts at in our emission right now, we only know half of our answer, which is that the final energy level should be at ultraviolet light since its highest energy. And we know an electron emission starts from a high energy level to a low energy level. So solving for initial energy level, we want to recall our equation where energy can be found from taking planks constant, multiplied by the speed of light, which is then divided by the wavelength of our radiation. So we're going to plug in our values to find our value for energy of our electron or rather of the photon of light for our radiation. And this will then lead us into solving using our reads berg equation, which we will get to later to find our initial energy level of the electron. So, plugging in what we know for at least the formula here, we can say that our energy of our photon is equal to recall that plank's constant H is represented by the value 6.626 times 10 to the negative 34th power with units of jewels, time seconds. And this is multiplied by our speed of light, which we should recall is a value of 3.0 times 10 to the eighth power units meters per second. So in our denominator we are dividing by the wavelength of our radiation which is actually given to us in the prompt in units of nanometers as 92.315 nanometers. However, we need our wavelength to be in units of meters. So we're going to multiply by a conversion factor in our denominator where we recall that our prefix nano tells us that we have 10 to the negative ninth power of our base unit meter. So this allows us to cancel out nanometers, leaving us with meters as our final unit in the denominator for wavelength and calculating everything out. We want to cancel out our units of meters. Now since we have it in the numerator and denominator as well as seconds, which is going to leave us with jewels as our final unit for our energy of our photon. And this is going to yield a value once we plug this into our calculators of 2.1533 times 10 to the negative 18th power jewels. And now that we know the energy of our photon of light of infrared light associated with our electrons. We want to find the change in energy to solve for initial energy levels. So we're going to recall our Rydberg equation where we have our Rights burg constant multiplied by -1, which is then multiplied by the difference between one over our final energy level squared subtracted from one over our initial energy level of our electron squared. So we want to plug in what we know into this formula to solve for the change in energy. Now we're sorry to sulfur initial energy level. We know that our energy change of our electron is equal to this value here. So we're going to plug that in. So we have 2.1533 times 10 to the negative 18th power jewels set equal to our right hand side, where we have negative one multiplied by our rights burg constant. So recall that that value of our constant is 2.18 times 10 to the negative 18th power with units of jewels. And then we want to take one over our final energy level squared. Which above we determined is that ultraviolet light since its highest in energy. So that would be over one squared, Subtracted from one over our initial energy level squared, which is what we are solving for. So to simplify this, we're going to divide both sides of our equation by our widespread constant 2.18 times 10 to the negative 18th power jewels, so that it cancels out on both sides. And sorry about that. So we're going to divide by that constant over here, 2.18 times 10 to the negative 18th power jewels. So let's get rid of the unit jewels as well. And what this will yield Is a value of 0.9877. Where we have that on the right hand side set equal to we can say 1 -1 over our initial energy level squared. So to get rid of that positive one, we're now going to subtract one from both sides of our equation. That's going to simplify to negative 0.0123 set equal to negative one over the initial energy level squared. Where not to get rid of the negative term. We're going to multiply both sides by -1. And that's going to happen on the left hand side where that will give us now positive 0.0123 equal to one over our initial energy level squared. And recall that when we have diagonals in algebra we can just switch places. So that's going to give us that our initial energy level squared is equal to a value of 81.3 where we will now take the square root of both sides to cancel out that square term. Where we can say that our initial energy level of our electron emission begins at a value of nine as the energy level. And because this is our initial energy level, it makes sense that this was an electron emission according to the prompt because we began at a higher energy level. So this is our second final answer And then we determined above that our first or sorry, final energy level for our first answer was equal to one at ultraviolet light. So everything highlighted in yellow represents our two final answers. I hope everything that I reviewed was clear if you have any questions leave them down below and I'll see everyone in the next practice video.

One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (b) Determine the initial and final values of n associated with this emission.

Verified Solution

Key Concepts

Hydrogen Emission Spectrum

Energy Level Transitions

Rydberg Formula