The Lyman series of emission lines of the hydrogen atom are
those...

Question

The Lyman series of emission lines of the hydrogen atom are
those for which nf = 1. (b) Calculate the wavelengths of the first three lines in the Lyman series—those for which ni = 2, 3, and 4.

Accepted Answer

Hey everyone in this example, we're told that for the lineman's theories are final energy level is the first energy level and we need to find the wavelength in nanometers of the emission lines for the initial energy levels 56 and seven. So we want to recall first our Rydberg equation which we represent as one over lambda. Our wavelength equal to our Rights Berg's constant, which is represented as our H multiplied by the difference between one over our initial energy. Or sorry, our final energy level squared minus one over our initial energy level squared. However, we want to reinterpret this so that we are solving for lambda for wavelength here. And so we should recall that we can also find wavelength by taking energy and setting that equal to Planck's constant times the speed of light and divided by our wavelength. So we would reinterpret our formulas here so that we have wavelength isolated and equal to our planks constant, multiplied by the speed of light and now divided by our rides Berg's constant, which is now going to have a negative value since we've moved it to the denominator and we're still going to have the multiplication of the difference between one over our final energy level squared minus one over our initial energy level squared. So beginning with the initial energy level five, we would calculate wavelength so that we have in our numerator plank's constant, which we recall is 6.626 times 10 to the negative 34th power jewels, time seconds. We would also plug in our speed of light, which we recall is three times 10 to the eighth power meters per second. And then in our denominator we want to recall our rights burg constant with a negative value is going to be 2.18 times 10 to the negative 18th power jewels. So now we want to multiply this by one over our final energy level squared which according to the prompt, our final energy level is the first energy level. So we would have won over one squared and this is going to be subtracted from one over our initial energy level, which for the first emission line is going to be five squared. So what we should recognize is because we placed our Rydberg constant from the numerator to now in the denominator we need to place this entire quotient over one. So what we would have is this entire quotient being placed over one. So we can go ahead and also fix this up here for our equation that we referenced. And what we're going to get is first by canceling out our units. We can cancel out jules. We can also cancel out seconds and we're left with meters for our wavelength value. So to simplify this, we would have that our wavelength is equal to on the left hand side. We should get a value equal to nine negative 9.12 times 10 to the eighth power And we're going to be multiplying this by 1/1 minus for 1/5 squared, you're going to get 0.4. And so when we take the value of these two products here and just a quick correction, this should be 10 to the negative eighth power. So back to taking the products of these two values here, we're going to get a value for wavelength equal to negative 9.5 times 10 to the negative eighth power. And our units are meters here. So we should have wrote meters there as well. Now we want wavelength to be in units of nanometers. So we should recall that to convert from meters to nanometers. We're going to recall that our prefix nano gives us 10 to the ninth power nanometers. So now we're able to cancel out units of meters leaving us with nanometers as our final unit for wavelength. And this is going to give us a value equal to negative 95 point oh nanometers as our wavelength. Now we want to recognize that the only reason why we are interpreting this answer as negative is due to the fact that this is an emission and for emissions, we emit our electrons so that is why we have this negative sign. However, we should interpret our wavelength by the positive value being 95 nanometers. But for the sake of our question here, we're going to leave this as our first answer. So this is our answer for the initial energy level being five. So moving on to the next line in our mission, we have the initial energy level being at six. So to simplify our solution, we already know what this portion of our equation equals. It equals this value here. So we're just going to carry that same value down so that we have wavelength is equal to negative 9.12 times 10 to the negative eighth power meters multiplied by. The only part that's going to change here is we would have won over 1 - Our fraction, which is now going to be 1/6 squared. And what we should get here is a value is a value equal to 9.38 times 10 to the negative eighth power. And we have units of meters still. So to get this value, you should just be able to type this entire product here in our calculators and we would get this value. But we again want our wavelength to be in nanometers. So we would convert from meters into nanometers by recalling that. We have 10 to the ninth nanometers in one m because our prefix nano means 10 to the ninth power. So now we're able to cancel out meters were left with nanometers as our final unit. And this is going to give us a wave length equal to negative 93.8 nanometers. So this would be our answer for the wavelength when we have the initial energy level in our mission being at six. And so that leaves us with our last emission, which according to the prompt is at The initial energy level of seven. So we're going to calculate that below. We would have an I equal to seven where we have the wavelength equal to our quotient value for Plank's constant times the speed of light divided by the negative rights rights, Berg's constant, which again was equal to negative 9.12 times 10 to the negative eighth power meters. And now we're multiplying this by 1/1 minus one over our initial energy level, which is now seven squared. This is going to give us a value equal to nine point or negative nine, sorry. And this value here should have also been negative. So it's negative 9.31 times 10 to the negative eighth power meters. And again we want wavelength to be in nanometers. So following our conversion from before we have 10 to the ninth nanometers in one m we would cancel out meters were left with nanometers. And for our last answer, we're going to get a result for a wavelength equal to negative 93.1 nanometers at the emission energy level being at seven. And so everything boxed in blue represents our final answers for our emission line 56 and seven for the initial values. So if you have any questions, please leave them down below. Otherwise, I hope I was able to help you understand and I'll see everyone in the next practice video.

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (b) Calculate the wavelengths of the first three lines in the Lyman series—those for which ni = 2, 3, and 4.

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