If each orbital can hold a maximum of `3` electrons, the number of elements in `4th` periodic table (long form) is.

Answer

To solve the question regarding the number of elements in the 4th periodic table (long form) given that each orbital can hold a maximum of 3 electrons, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Period**:
- The question specifies the 4th period of the periodic table. The principal quantum number \( n \) for the 4th period is 4.

2. **Determine the Types of Orbitals**:
- In the 4th period, the possible types of orbitals are:
- s-orbital (l = 0)
- p-orbital (l = 1)
- d-orbital (l = 2)
- f-orbital (l = 3)
- For \( n = 4 \), the orbitals available are:
- 1 s-orbital
- 3 p-orbitals
- 5 d-orbitals
- 7 f-orbitals (though f-orbitals start from the 6th period, we will only consider s, p, and d for the 4th period).

3. **Count the Number of Orbitals**:
- The total number of orbitals in the 4th period is:
- s: 1
- p: 3
- d: 5
- Total orbitals = \( 1 + 3 + 5 = 9 \) orbitals.

4. **Calculate the Maximum Number of Electrons**:
- Each orbital can hold a maximum of 3 electrons. Therefore, the total number of electrons that can be accommodated in the 4th period is:
\[ \text{Total electrons} = \text{Total orbitals} \times \text{Electrons per orbital} = 9 \times 3 = 27 \text{ electrons}. \]

5. **Determine the Number of Elements**:
- Each element corresponds to an electron. Thus, the number of elements in the 4th period is equal to the maximum number of electrons that can be accommodated, which is 27.

### Final Answer:
The number of elements in the 4th periodic table (long form) is **27**.

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