# What characteristic is given by the angular momentum quantum number?

##### 1 Answer

The **angular momentum quantum number**

Note that:

#l_max = n - 1# #l = 0, 1, 2, 3, . . . , n - 1# #2l + 1# is the number of orbitals in a given#l# subshell.#l# is the number of angular nodes in an orbital.

So two examples:

#1s# orbital:

#bb(n = 1)# for the principal quantum number (read it off the orbital name).Thus,

#l_max = bb(l = 0)# . This also means there isno angular nodein the#1s# orbital.

#2l + 1 = 2(0) + 1 = bb(1)# . Thus, there is onlyone#1s# orbital.That is also determined by realizing that for the

magnetic quantum number,#m_l = {0}# due to#l = 0# . i.e. the only#1s# orbital there is has#l = 0# .

#2s, 2p# orbital(s):

#bb(n = 2)# for the principal quantum number (read it off the orbital name).Thus,

#bb(l = 0, 1)# , and#l_max = 1# . This also means there isno angular nodein the#2s# orbital, but#bb(1)# angular nodein each#2p# orbital.

- For the
#2s# orbital(s),#2l + 1 = 2(0) + 1 = bb(1)# . Thus, there is onlyone#2s# orbital.That is also determined by realizing that for the

magnetic quantum number,#m_l = {0}# due to#l = 0# . i.e. the only#1s# orbital there is has#l = 0# .

- For the
#2p# orbital(s),#2l + 1 = 2(1) + 1 = bb(3)# . Thus, there arethree#2p# orbitals.That is also determined by realizing that for the

magnetic quantum number,#m_l = {-1, 0 +1}# due to#l = 1# . i.e. each#2p# orbitaluniquelycorresponds to either#l = -1# ,#0# , or#+1# . As there arethree#m_l# values, there arethree#2p# orbitals.