In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields ?
(A) n=1, l=0 m=0, (B) n=2, l=0, m=0
(c) n=2, l=1, m=1 (d) n=3, l=2, m=1
(e) n=3, l=2, m=0

Options

A. D and E
B. C and D
C. B and C
D. A and B

Explanation

To determine which orbitals have the same energy in a multi-electron atom, we can use the n + l rule. According to this rule, orbitals with the same value of n + l will have the same energy in the absence of magnetic and electric fields. Let's analyze each option step by step.

### Step-by-Step Solution:

1. **Option (A): n=1, l=0, m=0**
- Calculate n + l:
- n + l = 1 + 0 = 1

2. **Option (B): n=2, l=0, m=0**
- Calculate n + l:
- n + l = 2 + 0 = 2

3. **Option (C): n=2, l=1, m=1**
- Calculate n + l:
- n + l = 2 + 1 = 3

4. **Option (D): n=3, l=2, m=1**
- Calculate n + l:
- n + l = 3 + 2 = 5

5. **Option (E): n=3, l=2, m=0**
- Calculate n + l:
- n + l = 3 + 2 = 5

### Summary of n + l Values:
- Option (A): n + l = 1
- Option (B): n + l = 2
- Option (C): n + l = 3
- Option (D): n + l = 5
- Option (E): n + l = 5

### Conclusion:
From the calculations, we see that options (D) and (E) both have the same value of n + l (which is 5). Therefore, in the absence of magnetic and electric fields, these two orbitals will have the same energy.

### Final Answer:
**The orbitals that will have the same energy are (D) n=3, l=2, m=1 and (E) n=3, l=2, m=0.**

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