If the radius of inner most electronic orbit of a hydrogen atom is `5.3xx10^(-11)` m, then the radii of `n=2` orbit is

Options

A. `1.12 Å`
B. `2.12 Å`
C. `3.22 Å`
D. `4.54 Å`

Explanation

To find the radius of the n=2 orbit of a hydrogen atom when the radius of the innermost electronic orbit (n=1) is given as \(5.3 \times 10^{-11}\) m, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- The radius of the innermost orbit (n=1), denoted as \(R_1\), is \(5.3 \times 10^{-11}\) m.
- We need to find the radius of the n=2 orbit, denoted as \(R_2\).

2. **Use the formula for the radius of orbits in a hydrogen atom**:
- The formula to calculate the radius of the nth orbit is given by:
\[ R_n = n^2 \times R_1 \]
- Here, \(n\) is the principal quantum number of the orbit.

3. **Substitute the values into the formula**:
- For \(n=2\):
\[ R_2 = 2^2 \times R_1 \]
- Substituting \(R_1\):
\[ R_2 = 4 \times (5.3 \times 10^{-11} \text{ m}) \]

4. **Calculate \(R_2\)**:
- Performing the multiplication:
\[ R_2 = 4 \times 5.3 \times 10^{-11} \text{ m} = 21.2 \times 10^{-11} \text{ m} \]

5. **Convert to appropriate units**:
- To express \(R_2\) in angstroms (1 angstrom = \(10^{-10}\) m):
\[ R_2 = 21.2 \times 10^{-11} \text{ m} = 2.12 \text{ Å} \]

### Final Answer:
The radius of the n=2 orbit is \(2.12 \text{ Å}\).

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