"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i
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Answer:
The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;
(b) 4, 3, -3, 1/2.
Explanation:
To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;
4, 4, 4, 1/2
4, 3, -3, 1/2
4, 3, 0, 0
4, 5, 7, -1/2
4, 4, -5, 1/2
Let us label them as a to as follows
(a) 4, 4, 4, 1/2
(b) 4, 3, -3, 1/2
(c) 4, 3, 0, 0
(d) 4, 5, 7, -1/2
(e) 4, 4, -5, 1/2
Next we note the rules for the assignment and arrangement of quantum numbers are as follows
Number Symbol Possible values
Principal Quantum Number .......n........................1, 2, 3, ......n
Angular momentum quantum
number...............................................l.........................0, 1, 2, .......(n - 1)
Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l
Spin Quantum Number.................m.....................+1/2, -1/2
We are meant to analyze each of the arrangement for acceptability.
Therefore for (a),
we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.
Therefore (a) is not acceptable.
(b) Here we note that
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable
The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.
(c) Here we have
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable
The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 0 ∉ (+1/2, -1/2) → not acceptable
Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.
(d) Here we have;
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable
The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = -1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.
(e) Here we have;
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable
The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.
Remember your kinematic equations for constant acceleration. One of the equations is 
, where 
= final position, 
= initial position, 
= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means = 100m. You final position is where you ended up after t seconds passed, so = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
Your initial position is where you initially were before you braked. That means
Your acceleration is approximately