The electrons, identified by quantum number n and l
i. `n = 4,l=1` ii. `n = 4, l= 0` iii. `n = 3 , l= 2 ` iv. `n= 3 , l = 1`
Can be palced in the order of increasing energy from the lowest to highest,its

Options

A. `(iv)lt(ii)lt(iii)lt(i)`
B. `(ii)lt(iv)lt(i)`
C. `(i)lt(iii)lt(ii)lt(iv)`
D. `(iii)lt(i)lt(iv)lt(ii)`

Answer

To determine the order of increasing energy for the electrons identified by the given quantum numbers (n and l), we will use the N + l rule. This rule states that the energy of an electron in an atom increases with increasing values of the sum of the principal quantum number (n) and the azimuthal quantum number (l).

### Step-by-Step Solution:

1. **Calculate N + l for each set of quantum numbers:**
- For **i. n = 4, l = 1**:
\[ N + l = 4 + 1 = 5 \]
- For **ii. n = 4, l = 0**:
\[ N + l = 4 + 0 = 4 \]
- For **iii. n = 3, l = 2**:
\[ N + l = 3 + 2 = 5 \]
- For **iv. n = 3, l = 1**:
\[ N + l = 3 + 1 = 4 \]

2. **List the results:**
- i. \( N + l = 5 \)
- ii. \( N + l = 4 \)
- iii. \( N + l = 5 \)
- iv. \( N + l = 4 \)

3. **Order the values of N + l from lowest to highest:**
- The lowest value is from (ii) and (iv) which both equal 4.
- The next higher value is from (i) and (iii) which both equal 5.

4. **Determine the order based on N + l values:**
- Since (ii) and (iv) both have \( N + l = 4 \), we compare their n values:
- For (ii), \( n = 4 \)
- For (iv), \( n = 3 \)
- Since \( n = 3 \) is lower than \( n = 4 \), (iv) has lower energy than (ii).

5. **Final order from lowest to highest energy:**
- iv. \( n = 3, l = 1 \) (Lowest energy)
- ii. \( n = 4, l = 0 \)
- iii. \( n = 3, l = 2 \) (Same N + l as i but lower n)
- i. \( n = 4, l = 1 \) (Highest energy)

Thus, the final order of increasing energy is:
**iv < ii < iii < i**