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Problem 31 Hard Difficulty

Hydrogen's single electron can occupy any of the atom's distinct quantum states. Determine the number of distinct quantum states in the (a) $n=1,(\mathrm{b}) n=2,$ and $(\mathrm{c}) n=3$ energy levels.

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Top Physics 103 Educators
Andy C.

University of Michigan - Ann Arbor

Zachary M.

Hope College

Aspen F.

University of Sheffield

Meghan M.

McMaster University

Video Transcript

in this exercise, we have to determine how many different possible quantum states there are in different energy levels of the hydrogen. ETA in question one, you have to do that for an equals long. And remember that the the orbital control number l ranges from zero two and minus one. In our case here, the only allowed orbitofrontal number is zero. This means that necessarily the only allowed magnetic quantum number is also zero. So, up until now, we only have the We have an equals one l equals zero and m l equals zero. But we're also we also can have two possible speed Quintal numbers. Okay, you can have it. B plus 1/2 or plus or minus 1/2. Okay, so in total, there are two states for any clothes. OK, there there is the state l equals zero ml equals zero and M s equals close in half. And there's the state l equals zero m. L equals zero and m s minus one. Yeah, the question be, you have to do that for and equals two. So that case, the allowed orbits of quantum numbers are zero and one, so there are two allowed confront um, orbital quintal numbers. And for each allow or want to numbers, we have to determine how many allowed magnetic quantum numbers we have. So let's start for Al equals zero. In that case we have in l is equal to zero that there's only one possibility here. Okay, but a mess maybe plus 1/2 or minus 1/2. Okay, so there are two different states with l equals zero for l equals one. The allowed magnetic quintal number an M l equals two minus one ml. It was zero or ml. It was one. Okay, um, And for each m l, there are two possible spin quantum number. So blows when half our a minus 1/2. So there are also shoe for the amount from M. L equals zero two for m l equals one. So in total, there are six possible states for l one, so I'm gonna write it like this. There are six. State four. Well, it was one. There are two states for l equals zero. So in total, there are eight state four and equals two. Okay, in question. See, we have to do the same thing for enables. Three. So for an ID equals three the possible values of L R. Zero one or two. Okay, so let's start with l equals zero again. They're gonna be only two allowed states for the same reason that there there are only two allowed states in question be That's another change for l equals one. There's gonna be six allowed states again. Not gonna change. It's the same thing. Ah, it's the same line of thought. Ah, that that we developed for question be. And for l tables too, we have the falling possible values of M l. You can have ml It was mine to minus two minus one zero one or two. Okay, there are five possible ah state for l equals two. And for each ml there two possible states. Okay. Ah, uh, corresponding to the two states of M s of the speeding magnetic, it's being quantum number. So there is this beam speed plus when half and this being minus 1/2. So again, there are five possible values of M l for l equals two and there are two possible values of a mass for each ml. So in total there, 10 states for no equals two. Okay, 10 possible states. So, in total, we have two states for El zero, uh, six states for L one and 10 states for l two. In total, there are 18 states here and equals three, okay.

Universidade de Sao Paulo
Top Physics 103 Educators
Andy C.

University of Michigan - Ann Arbor

Zachary M.

Hope College

Aspen F.

University of Sheffield

Meghan M.

McMaster University