# How many possible states are there for an electron in a hydrogen atom when the principle quantum number (a) n= 1 and (b) n =2.?

##### 1 Answer

For a one-electron atom, the number of states is given by the number of quantum states available in all the orbitals that could be occupied. You can actually do this by hand without taking too much time.

Thus, there are

#ul(uarr color(white)(darr))# #+# spin-down possibility

#1s#

#ul(uarr color(white)(darr))# #+# spin-down possibility

#2s#

#ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))# #xx 3# for three orbitals

#underbrace(" "" "" "" "" "" "" "" ")#

#" "" "" "2p#

#ul(darr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))# #xx 3# for three orbitals

#underbrace(" "" "" "" "" "" "" "" ")#

#" "" "" "2p#

- At
#n = 1# , we have only the#1s# orbital,#l = 0# . - At
#n = 2# , we have the#2s# and#2p# orbitals,#l = 0,1# .

The **degeneracy**

#n = 1, l = 0 -> 2l + 1 = 1#

#n = 2, l = 0 -> 2l + 1 = 1#

#n = 2, l = 1 -> 2l + 1 = 3#

This gives

#ul(uarr color(white)(darr))# #+# spin-down possibility

#1s#

#ul(uarr color(white)(darr))# #+# spin-down possibility

#2s#

#ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))# #xx 3# for three orbitals

#underbrace(" "" "" "" "" "" "" "" ")#

#" "" "" "2p#

#ul(darr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))# #xx 3# for three orbitals

#underbrace(" "" "" "" "" "" "" "" ")#

#" "" "" "2p#