The fourth quantum number, known as the spin quantum number (s), can take on values of +1/2 or -1/2. For a 1s¹ electron, which is the only electron in the 1s orbital, the spin quantum number can be either +1/2 or -1/2, depending on the orientation of its spin. Therefore, the fourth quantum number for a 1s¹ electron could be either +1/2 or -1/2.
The quantum number that determines the size of an electron's orbit in a hydrogen atom is the principal quantum number, denoted by "n." For an electron orbit with a 31 Å diameter, the closest principal quantum number would be n = 4, because the average radius of the electron for an orbit corresponding to n is approximately given by n^2 Å in hydrogen atom.
I am checking the Wikipedia article on "quantum number", and don't find a quantum number "i" for the electron. If you mean "l", it seems that "l" can be between 0 and n-1. So, for n = 3, l can be between 0 and 2. If this is what you mean, I don't see any reason that would forbid this particular combination.
To determine the energy in the f-level orbit, you would first need to know the quantum numbers of the electron in that orbit, including the principal quantum number (n) and the azimuthal quantum number (l). The energy of an electron in a specific orbit is given by the formula E = -13.6 eV/n^2, where n is the principal quantum number. By plugging in the appropriate value of n for the f-level orbit (typically n = 3 or higher), you can calculate the energy of an electron in that orbit.
secondary quantum numberI don't think it is a number, but it could be referring to the Orbitals, being S, P, D, and F. Each orbital is a specific shape and the orbitals are determined in blocks on the Periodic Table. The energy, or Quantum Number would go in front, such as 4p, which means the principal quantum number or energy level is 4 and the orbital shape is p. Hope this helps
The magnetic quantum number, ml, runs from -l to +l (sorry this font is rubbish the letter l looks like a 1) where l is the azimuthal, angular momentum quantum number. The magnetic quantum number ml depends on the orbital angular momentum (azimuthal) quantum number, l, which in turn depends on the principal quantum number, n. The orbital angular momentum (azimuthal) quantum number, l, runs from 0 to (n-1) where n is the principal quantum number. l= 0 is an s orbital, l= 1 is a p subshell, l= 2 is a d subshell, l=3 is an f subshell. The magnetic quantum number, ml, runs from -l to +l (sorry this font is rubbish the letter l looks like a 1). ml "defines " the shape of the orbital and the number within the subshell. As an example for a d orbital (l=2), the values are -2, -1, 0, +1, +2, , so 5 d orbitals in total.
Ms = + 1/2
ms = -1/2
ms = +1/2
mi=0
The highest energy electron in a gallium atom would be found in the outermost shell, which is the fourth shell (n=4). The quantum number for this electron would be n=4, l=3, m_l=-3, m_s= +1/2.
mi=0
The principal quantum number n = 3 and the azimuthal or orbital angular momentum quantum number would be l =1 .l = 1
The first quantum number (n) represents the energy level (shell), so for a 1s2 electron, it would have a value of 1.
The last electron in gold is located in the 6s orbital. Therefore, the quantum numbers for this electron would be n=6 (principal quantum number), l=0 (azimuthal quantum number), ml=0 (magnetic quantum number), and ms=+1/2 (spin quantum number).
Either +1/2 or -1/2; the fourth quantum number is ALWAYS either +1/2 or -1/2 and it's not generally possible to say which (other than that two electrons in the same atom which have the same first three quantum numbers will always have different values for the fourth).
The first quantum number is the principal quantum number, denoted by "n." In aluminum, the 3p1 electron would have a principal quantum number of n = 3, since it is in the third energy level orbiting the nucleus.
The second quantum number refers to the azimuthal quantum number, also known as the angular momentum quantum number. For an electron in the 1s orbital of phosphorus (1s2), the azimuthal quantum number is 0, which corresponds to an s orbital. Therefore, for a 1s2 electron in phosphorus, the second quantum number would be 0.