Question

In the ground state Kr atom, how many electrons have both quantum number l = 2 and m_s= -1/2?What is the maximum amount of orbitals (in any atom) that could
have n = 3,   m_l= -1?

What is the maximum amount of electrons in any atom can have both of these quantum numbers?
n = 2 and m_l= 0?

In the ground state Br atom, how many electrons have quantum number l = 0?

In the ground state Xe atom, how many electrons have quantum number n = 4?_

What is the maximum number of electrons in an atom that can have both quantum numbers? n = 4, m_l = 1?

Answer 1

Recall Pauli Exclusion principle, which states that no two electrons can have the same quantum numbers. That is, each electron has a specific set of unique quantum numbers.

Now, let us define the quantum numbers:

n = principal quantum number, states the energy level of the electron. This is the principal electron shell. As n increases, the electron gets further and further away. "n" can only have positive integer numbers, such as 1,2,3,4,5,... Avoid negative integers, fractions, decimals and zero.

l =  Orbital Angular Momentum Quantum Number. This determines the "shape" of the orbital. This then makes the angular distribution. Typical values depend directly on "n" value. then l = n-1 always. Note that these must be then positive integers, avoid fractions, decimals. Since n can be 1, then l = 1-1 = 0 can have a zero value.

ml = Magnetic Quantum Number. States the orientation of the electron within the subshell. Therefore, it also depends directly on the "l" value. Note that orientation can be negative as well, the formula:

ml = +/- l values, therefore, 0,+/-1,+/- 2,+/-3 ... Avoid fractions and decimals

ms = the electron spin, note that each set can hold up to two electrons, therefore, we must state each spin (downwards/upwards). It can only have two values and does not depends on other values,

ms can cave only +1/2 or -1/2 spins. avoid all other numbers. also, avoid 0.5 or -0.5

In the ground state Kr atom, how many electrons have both quantum number l = 2 and ms= -1/2?

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

if l = 2, then, orbital is (0,1,2 ... s,p,d)

total "d" electrons --> 4p6 = 6 electrons

-1/2 spin --> only 50% of the 6 electrons

e- = 3 electrons will have same l = 2, -1/2

What is the maximum amount of orbitals (in any atom) that could have n = 3,   ml = -1?

if n = 3, then, l = n-1 = 3-1 = 2

if l = 2, then

l = 0 --> will not have ml = -1

l = 1, will have ml = -1

l = 2 --> will have ml = -1

therefore, 2 orbitals, p,d will have