An electron in an atom is known to be in a state with magnetic quantum number ml=0. What is the smallest possible value of the principal quantum number of the state?

1

Explanation:

The principal quantum number, n, showsthe principal electron shell. which can inturn be describe as the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as n=1 designates the first principal shell (the innermost shell). When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where n=2

As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on. n=1,2,3,4…

On the other hand, the magnetic quantum number ml determines the number of orbitals and their orientation within a subshell. Consequently, its value depends on the orbital angular momentum quantum number l. Given a certain l, ml is an interval ranging from –l to +l, so it can be zero, a negative integer, or a positive integer. ml=−l,(−l+1),(−l+2),…,−2,−1,0,1,2,…(l–1),(l–2),+l

note also that: The orbital angular momentum quantum number l determines the shape of an orbital, and therefore the angular distribution. The number of angular nodes is equal to the value of the angular momentum quantum number l. Each value of l indicates a specific s, p, d, f subshell (each unique in shape.) The value of l is dependent on the principal quantum number n. Unlike n, the value of l can be zero. It can also be a positive integer, but it cannot be larger than one less than the principal quantum number (n-1): l=0,1,2,3,4…,(n−1).

So the answer is 1

yea probably

..