In a set of experiments on a hypothetical oneelectron atom, you m...

Question

In a set of experiments on a hypothetical oneelectron atom, you measure the wavelengths of the photons emitted from transitions ending in the ground level (n = 1), as shown in the energy-level diagram in Fig. E39.27

. You also observe that it takes 17.50 eV to ionize this atom. (a) What is the energy of the atom in each of the levels (n = 1, n = 2, etc.) shown in the figure?

Accepted Answer

Welcome back, everyone. We are making observations about L I two plus L I two plus is a lithium atom with two of its electrons removed. Now, what we have here is we have this energy level diagram that shows the wavelengths of lights emitted when the electrons fall from an excited state down to the lowest energy state, which is of course, N equals one. Now, we are told that for the single valence electron of L I two plus, it has an ionization energy of 122.4 electron volts. And we are tasked with finding what is going to be the atoms first four energy levels otherwise known as electron shells. So how are we going to go about this? Well, what we can say is that the energy that we are looking for is going to be equal to planks constant times the speed of light divided by our photo wavelength here. Now, for each of our energy levels, the energy of the photon is going to be the energy of our initial level minus the energy of our final level. What we can go ahead and do here is we can calculate the energy of our photon just in general. And then we'll be able to find each of our levels here. So let's go ahead and do that first and that the energy of our photon according to this formula right here is equal to 6. times 10 to the negative 34th times three times 10 to the eighth, all divided by whatever wavelength we are given. What this allows us to say is that our energy for a photon is equal to one point 989 times to the negative 25th jules meters per wavelength. However, what we want to set up is we want to set up our energy for photon as being in electron volts meters. So what we will do is we will say that 1.989 times 10 to the negative 25th Jes meters over Lambda, we can multiply by one electron volt divided by 1.6 times 10 to the negative 19th Jews. So what this gives us is 1.24 times 10 to the negative six electron volt meters per wavelength. Wonderful. So now let's go through each of our levels and find our energy levels here and here's how we are going to do it. Let me scroll down here just a little bit we have. First, let's observe energy level one, the ionization energy for our upper state has an energy equal to zero. So what we can simply say is 122.4 electron volts, which is our ionization energy of our single valence electron plus E one is equal to zero. So what this gives us is that E one is equal to negative 1.22 or sorry, 122.4 electron volts. Let's move on to level two here. So we have our initial energy level energy level at energy two minus E one is equal to one 0.24 times 10 to the negative six all divided by the wavelength emitted from N equals two to N equals one. And as we can see, we're given that by this lambda of 13. nanometers or 13.5 times 10 to the negative ninth meters. So what does this give us? We have that E two minus negative 1.22 point four is equal to our right hand side of our equation. Now what our right hand side of our equation. Now, our left hand side of our equation gives us is 91.9 electron volts. So now solving for E two, what we get is negative 30.5 electron volts. Wonderful for the 3rd and 4th levels. I'm gonna go ahead and change color here. Let's go ahead and calculate our energy level. For E three. We have E three minus E one is equal to 1.24 times 10 to the negative six divided by our wavelength emitted going from energy level 3 to 1. What this gives us is E three minus negative 122.4 is equal to 1. times 10 to the negative six divided by this time, looking at this lambda right here, 11.4 nanometers or 11.4 times to the negative ninth meters. Now, what the right hand side of our equation gives us is one oh 8. electron volts. And finally solving for E three, what we get is negative 13.6 electron volts. Finally moving on to level four here, same procedure as before we have E four minus E one is equal to 1.24 times 10 to the negative six divided by our wavelength admitted going from level four to level one, which as we can see is our final unused lambda value. So what does this give us? We have the E four minus negative 122.4 is equal to 1.24 times 10 to the negative six divided by 10.81 nanometers or 10.81 times 10 to the negative ninth meters. Now, the right hand side of our equation is going to give us 114.7 electron bolts which then solving four E four gives us negative 7.69 electron volts. So now we have found our energy levels for 123 and four, which correspond to our final answer choice of c thank you all so much for watching. I hope this is for your help. We will see you all in the next one.

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