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 is no angular node in the#1s# orbital.
#2l + 1 = 2(0) + 1 = bb(1)# . Thus, there is only one#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 is no angular node in the#2s# orbital, but#bb(1)# angular node in each#2p# orbital.
- For the
#2s# orbital(s),#2l + 1 = 2(0) + 1 = bb(1)# . Thus, there is only one#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 are three#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# orbital uniquely corresponds to either#l = -1# ,#0# , or#+1# . As there are three#m_l# values, there are three#2p# orbitals.
