Like

Report

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) $n=5$ ?

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 $=2 n^{2} .$

You must be signed in to discuss.

Numerade Educator

University of Washington

Simon Fraser University

University of Winnipeg

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