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Problem 30 Easy Difficulty

How many sets of quantum numbers are possible for a hydrogen atom for which (a) $n=1,$ (b) $n=2,$ (c) $n=3$ , (d) $n=4,$ and (e) $n=5$ ?

Answer

By the above process, we see that in $n=1,$ there are 2 elements; in $n=2,$ there are $8 ;$ in $n=3$ , there are 18 elements and so on, which indicates that the formula for finding the electron $=2 n^{2} .$

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Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Farnaz M.

Simon Fraser University

Jared E.

University of Winnipeg

Video Transcript

the number of possible quantum states, which I just used. The member sign here That pound sign is equal to two times in squared. So for part A when N is equal to one, the number of states is equal to two times one squared, which is equal to two. Your part B you have in is equal to two. So the number of states is going to be equal to two times two squared where two squared is four times two is equal to eight. So the number of states is equal to eight on the for part C we have in is equal to three. So the number of states is gonna be two times three squared three squared is nine claims to is 18. So the number of states is 18. Making box it in is their solution for part C.

University of Kansas
Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Farnaz M.

Simon Fraser University

Jared E.

University of Winnipeg