Skip to main content
Ch.6 - Electronic Structure of Atoms

Chapter 6, Problem 89b

Consider a transition in which the electron of a hydrogen atom is excited from n = 1 to n = . (b) What is the wavelength of light that must be absorbed to accomplish this process?

Verified Solution
Video duration:
8m
This video solution was recommended by our tutors as helpful for the problem above.
1090
views
Was this helpful?

Video transcript

Hey everyone in this example, we have an electron from a hydrogen atom emitting a frequency of 4322 nm at the fifth energy level. And we need to determine the final energy level of this electron. So whenever we see the phrase determine the final energy level, we want to go ahead and recall that. We're going to have to utilize our formula for potential energy, which we recall is symbolized by delta e. And we're going to have to solve for N to get the answer to this problem. So we also want to recognize that because we're using the potential energy formula, we want to find our energy value for our electron at its current energy level in the 5th shell. So we're going to go ahead and recall that energy is equal to Planck's constant, multiplied by the speed of light divided by our given wavelength. And so we also want to recognize that our wavelength should be in units of meters, but it's given to us in units of nanometers here. So we're going to make a conversion so we can go ahead and get into the calculation for our current energy of our electron at the fifth energy level. So we should have that energy is equal to in our numerator. We recall that plank's constant is 6.626 times 10 to the negative 34th power jewels, time seconds. And then this is going to be multiplied by the speed of light, which we recall is three times 10 to the eighth power meters per second. Now in our denominator, we're going to plug in that given wavelength 4322 nanometers. And again we want to use the conversion factor to go from nanometers into meters. And so we want to recall that our prefix nano tells us that we should have 10 to the negative ninth power meters. And so now we're able to cancel out our nanometer units, leaving us with meters, which is the proper unit for our wavelength here. And now we can focus on finding this calculation in our calculators. But before we do so let's go ahead and cancel out the units. So we can go ahead and get rid of meters because it's in the numerator. And in the denominator down here we can also get rid of the second in the denominator and the second in the numerator over here and then we're just left with jewels as our unit for energy, which is what we want. And so in our calculators, this quotient is going to give us a value equal to four 5993 times 10 to the negative 20th power jewels. And so this is the energy of our electron at our current energy level Which is the 5th energy level and equals five. And so our next step is to implement our potential energy formula. So we should recall that our potential energy formula delta E. Is equal to the negative value of our rides berg's constant, multiplied by Z squared over and squared. And we should recall that Z is referring to the atomic number of our element that we're dealing with in this case hydrogen. And then we want to recall that N again is our energy level also known as our shell where are electron is located at. And so what we're going to do is begin this formula by representing delta E as the negative value of our Energy level that we're currently at for our for our electrons. So we calculated that to be 4. Times 10 to the negative 20th power jewels. And sorry, let's just make that meter there. And so we're going to go ahead and set this equal to The right hand side where we should have our Rydberg constant, which we recall is negative 2.18 times 10 to the negative 18th power jewels. And so because we want to find our final energy level, we're going to multiply this by the difference between our atomic number of our element, which in this case is hydrogen. So when we refer to our periodic tables, we find hydrogen as the first element and we see that it has atomic number one. So we can say that Z is equal to one. So we're going to take our atomic number one and place that over our final energy level, which is going to be n squared. And we're gonna subtract that from our Atomic number for hydrogen. one Over our current energy level which is going to be five squared. And that is due to the fact that the problem tells us and equals five. So now we can go ahead and close the parentheses off because we have plugged in that difference there between our energy levels. And we want to go ahead and simplify this as much as possible to solve for N. So our first step to solve this is to divide both sides by this quantity here, negative 2.18 Times 10 to the negative 18th power jewels on both sides. And so in order to make this simpler, we're just going to do the division by first in our calculators typing this portion here And in doing so, we should get a value equal to 2.10977. And then we're going to take our exponents and just subtract them. And so what we should have is this is going to be multiplied by 10 to the negative second power. And so this is set equal to the difference in our energy levels. So we have the final energy level one over and squared minus 1/5 squared. Our current energy level. Our next step is to further simplify this by taking the left hand side in our calculators and calculating that value. And so what we would have is that that's equal to 0.021, And this is set equal to our right hand side, One over, n squared -1 over. We can now say 25 on the right hand side because we know that five squared is 25. And so now we can actually turn that fraction and simplify this even further by turning that fraction to a decimal. So we should have 0.02, Set equal to our final energy level -0.04. And so to continue to isolate for N. We're going to add 0.04 to both sides of the equation so that it cancels out on the right hand side. And this will simplify to 0.061098 on the left hand side. And then we're setting that equal to our final energy level one over and squared. So now we want to get rid of that square term at this step. And so we're going to take the square root of both sides of our formula here. And what we should have is that this is equal to 0. and set equal to one over. And now that we've canceled out that square unit. So now we want to recognize that when we have diagonals when we're doing algebra we can just switch places. And so what we would have is N is equal to one over 0.24718. And so now we can just go ahead and Type this quotient into our calculators and this is going to tell us that N is equal to 4.04, which we can round to about four. And so this is going to be our final answer to complete this example as the energy level that our electron proceeds to once. It emits the frequency given in the problem. And so this is going to be our energy level of our electron and our hydrogen atom. So we're at the 4th energy level and this corresponds to choice A in our multiple choice up here. And so this will complete this example as our final answer. If you have any questions, please leave them down below and I will see everyone in the next practice video.
Related Practice
Textbook Question

The watt is the derived SI unit of power, the measure of energy per unit time: 1 W = 1 J>s. A semiconductor laser in a DVD player has an output wavelength of 650 nm and a power level of 5.0 mW. How many photons strike the DVD surface during the playing of a DVD 90 minutes in length?

2645
views
Textbook Question

In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as a function of the frequency of radiation hitting a metal surface. She obtains the following plot. The point labeled 'n0' corresponds to light with a wavelength of 542 nm. (a) What is the value of n0 in s - 1?

625
views
Textbook Question

Consider a transition in which the electron of a hydrogen atom is excited from n = 1 to n = . (a) What is the end result of this transition?

764
views
Textbook Question

Consider a transition in which the electron of a hydrogen atom is excited from n = 1 to n = . (d) How are the results of parts (b) and (c) related to the plot shown in Exercise 6.88?

558
views
Textbook Question

The human retina has three types of receptor cones, each sensitive to a different range of wavelengths of visible light, as shown in this figure (the colors are merely to differentiate the three curves from one another; they do not indicate the actual colors represented by each curve):

(c) Explain why the sky appears blue even though all wavelengths of solar light are scattered by the atmosphere.

548
views
Textbook Question

The series of emission lines of the hydrogen atom for which nf = 3 is called the Paschen series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Paschen series are observed.

757
views