The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed.

Video transcript

Welcome back everyone in this example, We're asked to determine the region of the electromagnetic spectrum of the Balmer series. Seeing if its final energy state is at the second energy level. So we're going to recall that a bomber series describes an electron going from a energy level that is greater than the second energy level down to the second energy level. And this is creating a electron emission. Where on our electromagnetic spectrum we would have a bomber spectral line which describes our electron emission going from a higher or higher energy level to the second energy level. And we would recall that this would be visible to us on our electromagnetic spectrum, from the 380 nanometer wavelength to our 700 nanometer wavelength. So this would be what we recall as are visible wavelength spectrum. So even though we've recalled that these wavelengths of our Balmer series should occur in our visible region of our electromagnetic spectrum. We want to prove this with our following equation where we can recall that to calculate wavelength which we should recall is represented by lambda. We are going to set that equal to our product of plank's constant, multiplied by our speed of light. And then this is all divided by our energy of a photon. Now, for the energy of our photon, we want to recall that we're going to be multiplying it by the difference of one over our final energy level squared, subtracted from one over our initial energy level of our electron squared. So this should be continued and it must be final minus initial for our setup here. So we do know our final energy level is the second energy level. Since this is a bomber series Now for the initial energy level, we don't know that. But because we know it has to be greater than two, we can say that therefore the initial energy level is going to be equal because it must be greater than two. It's therefore going to be equal to about. I'm sorry. This should say an an initial here. And we labeled this as an F. So because we know it must be greater than two. We can just assume that it's going to be the third energy level. Or we can even say for or the fifth energy level, but we're just going to go with the third energy level. Or actually let's just go with the fourth energy level. So what we would have going into our calculation is that our wavelength is going to equal in our numerator. We would recall that plank's constant is a value of 6.626 times 10 to the negative 34th power with units of jewels times seconds. And then we would recall that our speed of light is equal to a value of three times 10 to the eighth power meters per second. As our units In our denominator, we would recall that the energy of a photon of light is equal to the value negative 2.18 times 10 to the negative 18th power units of jewels. And as we stated, we're going to be multiplying it by the difference here, we have one over which one represents our atomic number of hydrogen for a photon. And in our denominator we have our final energy level squared, which we agreed because of the Balmer series is two squared subtracted from one over our initial energy level squared. And because we're assuming that it's the fourth energy level, we would have four squared. And so in our calculators, we're going to carefully type this all out. So this is going to yield us a value where our wavelength is equal to. And sorry about that. Let's scroll back up. So our wavelength is going to equal a value of 4. and we'll say 0.86 times 10 to the negative seventh power. And as far as our units will be able to cancel out jewels with jewels in the denominator will be able to cancel out seconds. And that leaves us with meters as our final unit. But as we stated above, we recall that wavelength is expressed in nanometers. And so we're going to multiply by our conversion factor where we recall that to go from meters to nanometers are prefix nano tells us that we have for one m 10 to the ninth power nanometers. And sorry, this is a negative value here. So when we multiply, we're going to get a wavelength equal to negative 486 nanometers as our final unit here because meters will cancel out. And this would be our wavelength. However, we recall that wavelength is reported as a positive value. And so we're only getting a negative value here because this is an electron emission. And so we recall that wavelength is positive. And so therefore, we would say our wavelength is equal to 486 nanometers. And based on what we said is in the visible light spectrum, we recall that that's between any wavelength of 380 nanometers to nanometers. And so this is in the visible light spectrum. So this is in the visible region of the electrode magnetic spectrum. So, for our final answer, we will highlight our calculated wavelength which we confirmed is in the visible region of our electromagnetic spectrum. And we proved this above with our calculated wavelength at our assumed initial energy level at the fourth energy level for our bomber series. So this corresponds to choice C as the correct choice. I hope everything was clear. But if you have any questions, leave them down below. And I'll see everyone in the next practice video