Question

An atom has an electron with $m_l = 2$ . What is the smallest value of $n$ for this electron?

Final Answer

3

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 30, Problem 36 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We are told that the angular momentum projection number is 2 and we are asked to figure out what is the smallest possible value for the principal quantum number

*n*? Now we know that*n*is related to the angular momentum quantum number by this sequence here*l*goes from 0 up to*n*minus 1 in integers or whole numbers I guess you should say and so if we can minimize*l*then we'll minimize*n*and so we look at this sequence here for the possible values for the angular momentum projection number to figure out what the smallest possible value for*l*is. So we know*m l*is 2—that's for sure— now*l*can be 2, it cannot be 1 because the maximum possible value for*m l*is*l*and so given that*m l*is 2;*l*cannot be 1 since that would be less than 2;*l*could be 3 in which case 2 is allowed for*m l*because you can have 3 minus 1 to make 2 but we have to minimize*l*with the smallest possible 1 and the smallest possible number for*l*is 2 because that works because*m l*can be*l*itself. Okay! So 2 is the smallest value for the angular momentum quantum number. So the angular momentum quantum number goes from 0 all the way up to*n*minus 1 and so we can say*n*minus 1 at its smallest is going to be 2 in which case the smallest principal quantum number then is 3.