For the H₂ molecule the equilibrium spacing of the ...

Question

For the H₂ molecule the equilibrium spacing of the two protons is $0.074$ nm. The mass of a hydrogen atom is $1.67 \times 10^{- 27}$ kg. Calculate the wavelength of the photon emitted in the rotational transition $l equals 2$ to $l equals 1$ .

Accepted Answer

Hey everyone. So this problem is dealing with the atomic structure. Let's see what it's asking us. The spacing between two nuclei of a carbon monoxide molecule is 0.1135 nanometers consider the mass of the carbon atom as 1.993 times to the negative 26 kg. And that the oxygen atom is 2.656 times 10 to the negative 26 kg. A transition from L equals three to L equals two results in the emission of a photon. And we're asked to determine its wavelength. Our multiple choice answers are a 1. micrometers B 2.276 micrometers C 8.733 micrometers or D 873.3 micrometers. So the key to solving this problem is at first recognizing that the change in energy between levels is equal to the energy of the photon. And we so we can recall the equation for the change in energy between levels as well as the relationship between energy and a photon. And will come up with the two equations needed to solve this problem. So E sub L or the energy for a level is equal to L multiplied by the quantity L plus one multiplied by H bar squared, all divided by two I. And our energy in terms of a wavelength is delta E is equal to HC divided by lambda where lambda is the wavelength that we are trying to solve for. And so when we look at our change in energy between levels, we can look at the energy of level three minus the energy of level two for this problem. And so when we plug that in, we get three multiplied by 4h squared hr squared divided by two I minus for level two, two multiplied by three multiplied by H bar squared divided by two I. And that simplifies too three H bar squared divided by I. At the same time, we can recall that H bar or the reduced planks constant is just another way to write H divided by two pi. And so we can rewrite this as three H squared divided by four pi squared multiplied by I. And now we can set this delta E equation equal to our relationship between delta E and Lambda formula. And so that looks like three H squared divided by four pi squared I is equal to HC divided by LAMBDA. Because we're solving ultimately for LAMBDA, we're going to isolate that on the left hand side of the equation, recognize that one of the HS cancel from each side. And we're left with four pi squared C I all divided by three H. When we look at these terms C is a constant speed of light, H, constant planks constant. But we need to solve for I our intensity before we can solve for our ultimate goal of wavelength. And so the next step here is recalling our equation for intensity between two atoms. So we have I is equal to M sub R multiplied by R not squared where MS of R is our reduced mass. And that's given by the equation, the, each the mass of each atom multiplied together divided by the sum of those masses. And so for this, we have the mass of the carbon multiplied by the mass of the oxygen divided by the mass of the carbon plus the mass of the oxygen. And are, are not, is the distance between the two atoms. And that was given to us in the problem as 0.1135 nanometers which we can rewrite as 10 to the negative 9 m. And so the mass of the carbon was given in the problem as 1.993 times 10 to the negative 26 kg. The mass of the oxygen was given as 2.656 times 10 to the negative kg. And so now we have everything we need to solve for our intensity, we just have to plug it in. So we'll take the mass of the carbon one point times 10 to the negative 26 kg. Multiplied by the mass of the oxygen 2.656 times 10 to the negative 26 kg all divided by the mass of the carbon. Again 1.993 times 10 to the negative kg plus the mass of the oxygen 2.656 times 10 to the negative kg. And then we'll multiply that by just make sure we keep everything separated here. We'll multiply that by our um radius, which is that distance between the two atoms. So that will be 0.1135 times 10 to the negative 9 m. And that quantity is squared. And so we have now when we plug that in 1.4 6 6 kg meters squared, I'm sorry, 1.4668 times 10 to the negative 46. There we go kilogram meters squared. Now we're finally ready to solve four, our lambda. And so plugging in those values into that equation, we have four pi squared multiplied by C the speed of light, which we can recall is 2.998 times 10 to the m per second multiplied by R I which we just saw for 1.4668 times 10 to the negative 46 kg meters squared, all divided by three H three multiplied by our planks constant, which we can recall is 6. times 10 to the negative jeel seconds. And we plug that into our calculator and we get 8.733 times 10 to the negative 4 m. This is where units are really important. So all of the answers were given in units of micrometers which is 10 to the negative 6 m. So we can rewrite this as 873. micrometers. And so that is the final answer for this problem. When we go back up to our multiple choice answers, it aligns with answer choice D D is the correct answer for this one. That's all we have. We'll see you in the next video.

For the H₂ molecule the equilibrium spacing of the two protons is $0.074$ nm. The mass of a hydrogen atom is $1.67 \times 10^{- 27}$ kg. Calculate the wavelength of the photon emitted in the rotational transition $l equals 2$ to $l equals 1$ .

Key Concepts

Rotational Transitions

Energy of a Photon

Moment of Inertia

Watch next

For the H2 molecule the equilibrium spacing of the two protons is 0.0740.074 nm. The mass of a hydrogen atom is 1.67×10−271.67\times10^{-27} kg. Calculate the wavelength of the photon emitted in the rotational transition l=2 l = 2 to l=1l = 1.

Key Concepts

Rotational Transitions

Energy of a Photon

Moment of Inertia

Watch next

For the H₂ molecule the equilibrium spacing of the two protons is $0.074$ nm. The mass of a hydrogen atom is $1.67 \times 10^{- 27}$ kg. Calculate the wavelength of the photon emitted in the rotational transition $l equals 2$ to $l equals 1$ .