Asked by jazz
The electron in a hydrogen atom with an energy of -0.544 eV is in a subshell with 18 states.
What is the principal quantum number, n, for this atom?
What is the maximum possible orbital angular momentum this atom can have?
Is the number of states in the subshell with the next lowest value of l equal to 16, 14, or 12?
What is the principal quantum number, n, for this atom?
What is the maximum possible orbital angular momentum this atom can have?
Is the number of states in the subshell with the next lowest value of l equal to 16, 14, or 12?
Answers
Answered by
drwls
Bassed upon the energy level, n^2 = 13.6/.544 = 25, and n = 5. That n state has 2 n^2 = 50 substates.
The maximum orbital angular momentum quantum number for this n-value is
(lower case L) = n-1 = 4. The orbital angular momentum of that state is 4 h, where h is Planck's constant.
For the next lower orbital quantum number of that n-state, L = 3, there are
2*(2L + 1) = 14 subshells.
The L projections are -3, -2, -1, 0, 1, 2 and 3, with two electron spin projections per L projection.
The maximum orbital angular momentum quantum number for this n-value is
(lower case L) = n-1 = 4. The orbital angular momentum of that state is 4 h, where h is Planck's constant.
For the next lower orbital quantum number of that n-state, L = 3, there are
2*(2L + 1) = 14 subshells.
The L projections are -3, -2, -1, 0, 1, 2 and 3, with two electron spin projections per L projection.
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