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Apply the Pauli exclusion principle to determine the number of electrons that could occupy the quantum states described by $(\mathrm{a}) n=3, \ell=2, m_{\ell}=-1$ and (b) $n=3, \ell=1,$ and

(c) $n=4$.

(a) 2; (b) 6; (c) 32

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in question. A. We have to determine how many elections can occupy quantum states that have quantum numbers and equals three. L equals two and ml equals minus one. And in order to do that, we have to use policies, exclusion principle And what the what policies exclusion principle tells us is that states can only be a compiled by one electron at a time, so there can be more than one electron per state. So all we need to do this to determine how many states are possible given these quantum numbers. Ah, so we have the the principal Quantum member, the orbital Quintal number and the magnetic quantum numbered all are fixed. And the only thing that we can change here that, uh, that is allowed to change is the spin quantum number, so it can be either minus 1/2 or plus 1/2. So in the end, there are two possible states given these quantum numbers. So there are two elections. Okay, two electrons can occupy, uh, quantum states. With these numbers, the question be, you have to do the same thing now with the quantum numbers and equals three and l equals one, okay. And All we have to do now is to determine how many states there are with these windle numbers. So for l equals one and l can be either in the minus 10 or one. There are three possibilities and for each possibility, off an L. There are two possibilities for M s, which is minus which is plus, uh, sharp blows 1/2 or minus 1/2. So what we have to do is to multiply three by 23 possible values of the Mau two possible values of M s. So there are six electrons six electrons can occupy states that I have quantum numbers and equals three and our equals one in question. See, we have to answer the same question, but with an equals four. So for n equals four, we have yeah, values of El There ranged from zero to n minus one. So 0 to 3012 and three. Therefore, values off the orbital Quintal number, and for each orbit open to number, we can find how many states are allowed, and then we send them all. Okay, so for l equals zero, m l is necessarily zero. So there's only one magnetic quantum number allowed, but there are two m s is So it's a mess. Minus 1/2 for plus 1/2. So there are two allowed states for zero. Okay, there. I'm gonna write it in green. There to a lot of states for al equals one, we have and l maybe minus 10 or one. Okay. And again, there are two possible states were a mess minus 1/2 or pleasure in half. So in total, we have to mow supply three by two. So there are six possible states for l equals two. We have an L. It may be minus two minus 101 or two. So there are five m els there as always to possible spin states. So in the end, there are 10 possible states for Alec was too for l equals three. You have an l ranging for a minus three, 23 Okay, so there are seven possible. There are seven possible states. I'm sorry. There seven possible numbers and l. So, in total, we have to multiply seven by two. Because there are, as always, to possible spin numbers. So in total, there are 14 so, in order to know all of the possible the the number of boss possible states in these, the for the quantum number and equals four we have to send them. Also, it's two plus six plus plus 10 plus 14. So in total there are 32. There can be 32 electrons occupying states with mental number and equals four.

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