The decrerasing order of energy for the electrons represented by the following sets of quantum number is :
1.`n = 4,l = 0,m = 0,s = +- 1//2`
2.`n = 3,l = 1,m = 1,s = - 1//2`
3.`n = 3,l = 2,m = 0,s = + 1//2`
4.`n = 3,l = 0,m = 0,s = - 1//2`

Options

A. `1 gt 2 gt 3 gt 4`
B. `2 gt 1 gt 3 gt 4`
C. `3 gt 1 gt 2 gt 4`
D. `4 gt 3 gt 2 gt 1`

Answer

To determine the decreasing order of energy for the electrons represented by the given sets of quantum numbers, we will use the n + l rule. This rule states that the energy of an electron in an atom is primarily determined by the sum of the principal quantum number (n) and the azimuthal quantum number (l). The lower the value of n + l, the lower the energy level. If two electrons have the same n + l value, the one with the lower n value has lower energy.

Let's analyze each set of quantum numbers step-by-step:

1. **Set 1: n = 4, l = 0**
- Calculate n + l:
- n + l = 4 + 0 = 4

2. **Set 2: n = 3, l = 1**
- Calculate n + l:
- n + l = 3 + 1 = 4

3. **Set 3: n = 3, l = 2**
- Calculate n + l:
- n + l = 3 + 2 = 5

4. **Set 4: n = 3, l = 0**
- Calculate n + l:
- n + l = 3 + 0 = 3

Now, we have the following n + l values:
- Set 1: n + l = 4
- Set 2: n + l = 4
- Set 3: n + l = 5
- Set 4: n + l = 3

Next, we will arrange these sets based on the n + l values:
- The lowest n + l value is from Set 4 (3), which has the lowest energy.
- The next lowest n + l value is from Sets 1 and 2 (both 4). Since Set 2 has a lower n (3) than Set 1 (4), Set 2 has lower energy than Set 1.
- The highest n + l value is from Set 3 (5), which has the highest energy.

Thus, the decreasing order of energy for the electrons represented by the given sets of quantum numbers is:
1. Set 4 (n = 3, l = 0) - Lowest energy
2. Set 2 (n = 3, l = 1)
3. Set 1 (n = 4, l = 0)
4. Set 3 (n = 3, l = 2) - Highest energy

**Final Order: 4, 2, 1, 3**