Bilangan kuantum berikut yang tidak mungkin terjadi kedudukan elektron adalah __ a n=3,l=0,m=0,s=-1/2 b. n=3,l=1,m=0,s=+1/2 C. n=3,l=2,m=-3,s=+1/2 d. n=4,l=2,m=0,s=+1/2 e n=3,l=2,m=+2,s=+1/2

c. n=3, l=2, m=-3, s=+1/2

Here the question is about quantum numbers which are used to describe distinctly the state of a particle in Quantum mechanics. There are four quantum numbers: principal quantum number (n), the azimuthal quantum number (l), the magnetic quantum number (m), and the spin quantum number (s). The principal quantum number, n, consists of positive integer, determining the energy level and the average distance of electron from the nucleus. It can take any value starting from 1, as n = 1, 2, 3, ..The azimuthal quantum number, l, defines the shape of the orbital, and can take on the values, from 0 to n – 1 (including 0 and n – 1).The magnetic quantum number, m, defines the orientation of the orbital within the field, and ranges from -l to +l, including zero. And finally, the Spin quantum number s, which can only have two values +1/2 and -1/2 denoting the direction of spin as either up or down. For any given value of n, if l is greater than n-1 or m is larger than l or smaller than -l, it would mean the quantum state is in plausibility, which makes the quantum situation impossible.So, evaluating all options:a. The quantum state is plausible as for n = 3, l=0 are valid and m=0 falls under the condition m should be -l