The principal quantum number of H-atom orbital, if the energy of `e^(-)` is `-3.4eV`, will be:

Options

A. 1
B. 2
C. 3
D. zero

Explanation

To find the principal quantum number (n) of a hydrogen atom orbital when the energy of the electron is given as -3.4 eV, we can follow these steps:

### Step 1: Understand the formula for energy in a hydrogen atom
The energy of an electron in a hydrogen atom is given by the formula:
\[ E = -\frac{13.6 \, Z^2}{n^2} \]
where:
- \( E \) is the energy of the electron,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number.

### Step 2: Substitute the known values into the formula
Given that the energy \( E = -3.4 \, \text{eV} \) and for hydrogen \( Z = 1 \), we can substitute these values into the energy formula:
\[ -3.4 = -\frac{13.6 \times 1^2}{n^2} \]

### Step 3: Eliminate the negative sign
We can eliminate the negative signs from both sides of the equation:
\[ 3.4 = \frac{13.6}{n^2} \]

### Step 4: Rearrange the equation to solve for \( n^2 \)
Now, we can rearrange the equation to isolate \( n^2 \):
\[ n^2 = \frac{13.6}{3.4} \]

### Step 5: Calculate \( n^2 \)
Now, we perform the division:
\[ n^2 = \frac{13.6}{3.4} = 4 \]

### Step 6: Find the value of \( n \)
To find \( n \), we take the square root of both sides:
\[ n = \sqrt{4} = 2 \]

### Conclusion
The principal quantum number \( n \) for the hydrogen atom orbital, given the energy of the electron is -3.4 eV, is:
\[ n = 2 \]

### Final Answer
**The principal quantum number \( n \) is 2.**

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