The four quantum numbers of the valence electron of potassium atom are:
4, 0, 0, 1/2
Quantum numbers are a set of four numbers that describe the unique state of an electron in an atom. They provide information about the electron's energy level, shape of the orbital, orientation of the orbital in space, and its spin.
The question asks for the four quantum numbers of the valence electron of a potassium atom. To find this, we first need to determine the electronic configuration of potassium.
Potassium (K) has an atomic number of 19, meaning it has 19 electrons. The electronic configuration follows the Aufbau principle, Hund's rule, and the Pauli exclusion principle:
The electronic configuration of potassium is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$.
The valence electron is the electron in the outermost shell (the shell with the highest principal quantum number). In the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$, the outermost shell is the $n=4$ shell, which contains one electron in the $4s$ orbital. Therefore, the valence electron is the electron in the $4s^1$ orbital.
Now, let's determine the four quantum numbers for the valence electron in the $4s^1$ orbital:
For the valence electron of the potassium atom in the $4s^1$ orbital, the four quantum numbers are:
Therefore, the four quantum numbers are (4, 0, 0, 1/2).
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The quantum numbers n and l for four electrons are given below.
(i) n = 4, I = 1
(ii) n = 4, l = 0
(iii) n = 3, l = 2
(iv) n = 3, l = 1
The order of their energy from lowest to highest is: