Consider the hydrogen atom. A state of the hydrogen atom is specified by the qua
ID: 1502505 • Letter: C
Question
Consider the hydrogen atom. A state of the hydrogen atom is specified by the quantum numbers n, l and m.
(a) Given a value n for the principal quantum number, what are the allowed values of the orbital angular momentum quantum number l?
(b) Given a value l for the orbital angular momentum quantum number l, what are the allowed values of the magnetic quantum number m? What is the number of allowed values for m?
(c) Given the value n = 3 for the principal quantum number, show that the total number of possible states (i.e. combinations of values for the l and m quantum numbers) is equal to 9. (d) For any given value n for the principal quantum number, show that the number of possible states is equal to n^2
Explanation / Answer
If n=3 then l= 0,1,2 and m=-2,-1,0,1,2 this means that for l=0 you have m= (1 combination) for l 1 you have m=-1,0,1 (3 combinations) and 5 combinations for l equal 2, 1+3+5=9
Principal Quantum Number (n): n = 1, 2, 3, …,
Specifies the energy of an electron and the size of the orbital (the distance from the nucleus of the peak in a radial probability distribution plot). All orbitals that have the same value of n are said to be in the same shell (level). For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state. The total number of orbitals for a given n value is n2.