A small first-aid kit is dropped by a rock climber who is descending steadily at 1.3 m/s. After 2.5 s, what is the velocity of t
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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.
Answer:
(a). The magnitude of the uniform magnetic field is 10.02 μT
(b). The the magnitude of the total magnetic field is 10.78 μT.
Explanation:
Given that,
Number of turns = 210
Radius = 5 cm
Current = 38 mA
Current in the wire = 500 mA
We need to calculate the magnetic field inside a coil
Using formula of magnetic field
Put the value into the formula
Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil
Distance
We need to calculate the magnetic field from the straight wire
Using formula of magnetic field
Put the value into the formula
This field is perpendicular to the wire.
The magnitude of magnetic field is
Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT
(b). The the magnitude of the total magnetic field is 10.78 μT.
Answer:
1.36
Explanation:
= Index of refraction of air = 1
= Index of refraction of plastic = ?
i = angle of incidence in air = 32.0° deg
r = angle of refraction in plastic = 23.0° deg
Using Snell's law
Sini =
Sinr
(1) SIn32.0 = Sin23.0
= 1.36