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Two electrons in the same atom have $n=3$ and $\ell=1 .$ (a) List the quantum numbers for the possible states of the atom. (b) How many states would be possible if the exclusion principle did not apply to the atom?

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University of Sheffield

In this exercise, we have two electrons in the n equals three l equals wants of show. And in the first question, we have to list all possible quantum states of these two elections taking to account Polly's exclusion principle. But Pauli's exclusion principle tells us is that two electrons cannot be the same quantum states simultaneously. And what we need to remember is that the allowed values for the magnetic quantum number and l range from minus l so tell which in our case, are just minus 10 and one and the allowed values for the spin quantum number M s r plus or minus 1/2. So let's start listing all of these states. Ah, so I'm gonna do a little table, okay? It's not gonna be so little. Ah, And in this first table, I'm gonna fix that. The magnetic quantum number of the first electron is minus one. The spin quantum number of the first election is minus one, and I'm gonna list all possible states for the second elected. OK, so we have ML It was minus one m s equals plus one noticed that M s cannot be minus one because of their wise be the second election would be in the same state of the 1st 1 We just write this a little better then the other states are also M l equals zero m s equals, uh, plus or minus when half that's allowed M l equals zero. I'm sorry, m l equals one or a mess equals plus or minus 1/2. That's also a lot. Okay, then we have the second table where I'm going to fix that, and l equals minus one and ass. It was plus Ah, 1/2. So this is an l, uh, for the second, uh, the second electorally could have m l equals minus one. M s equals minus 1/2. It cannot be blessed with half for the same reason as before, and all the other states are allowed. So here, Okay. And I'm gonna keep, uh, making the these lists until all the states have enlisted. So m l zero. And as it was minus 1/2. Then the allowed r l. It was minus one. And as it was, plus or minus when half and l equals zero a mass. It was plus 1/2. There's all minus here. Bless 1/2 in the mail equals one. A mass equals plus or minus 1/2. Again, I have m l equals zero m s. It was plus 1/2. So they allowed our email. It was minus one m s equals, Uh, my pleasure, miners 1/2 and l equals zero M s equals minus 1/2 m l equals one m s equals plus or minus 1/2. And now for the last two tables. So for the first electron and the in this table, I'm gonna fix and l equals one. A mess equals minus 1/2 so and l Because minus one a mass equals plus or minus half a mass equals zero a mess. Mr. Email here equals zero. A massive gal's plus or minus 1/2 and ml. It was one a mass. It was closed from half. And finally, the last table we have. We I'm gonna fix em out, equals one. A mass equals plus 1/2. So I have m l equals minus one m s who's plus or minus 1/2 and l equals zero m s. It was Let's remind us 1/2 and and now it was one a Mass equal minus one had okay, These 30 states that I have just written. Okay, Are the allowed quantum states in question be? We have to calculate how many states would be possible if if it were not for the powerless export for Polish exclusion principle. If it didn't exist, how many states would be a lot in the case? As I said in question, a a male can be minus 10 or one, and m s can be either miners 1/2 or 1/2 blessed one. Has eso noticed that for each electric there are six possible states to uh So for the two elections, there would be six times six possible states possible combinations of state. Okay, so in total, there would be 36 states. Okay, there are actually fewer because of the part because of Paula's explosion principle. But, uh, if it were not for for for it, we would have 36 possible states for the two electrons

Universidade de Sao Paulo