The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with:

Options

A. Lyman series
B. Balmer series
C. Paschen series
D. Brackett series

Explanation

To solve the problem, we need to determine which series the transition of an electron in a hydrogen atom corresponds to, given that it ends up in an orbital with a radius of 211.6 pm (picometers).

### Step-by-Step Solution:

1. **Understand the Formula for Orbital Radius**:
According to Bohr's model of the hydrogen atom, the radius of the nth orbital is given by the formula:
\[ R_n = 0.529 \times \frac{n^2}{Z} \text{ angstroms} \]
where \( R_n \) is the radius of the nth orbital, \( n \) is the principal quantum number, and \( Z \) is the atomic number (which is 1 for hydrogen).

2. **Convert the Radius**:
The radius given in the question is 211.6 pm. We need to convert this to angstroms for consistency:
\[ 1 \text{ pm} = 10^{-12} \text{ m} \quad \text{and} \quad 1 \text{ angstrom} = 10^{-10} \text{ m} \]
Therefore,
\[ 211.6 \text{ pm} = 211.6 \times 10^{-12} \text{ m} = 2.116 \times 10^{-10} \text{ m} = 2.116 \text{ angstroms} \]

3. **Set Up the Equation**:
Now we can set up the equation using the radius formula:
\[ 2.116 = 0.529 \times \frac{n^2}{1} \]
Simplifying this gives:
\[ n^2 = \frac{2.116}{0.529} \]

4. **Calculate \( n^2 \)**:
Performing the division:
\[ n^2 \approx 4 \]
Taking the square root gives:
\[ n = 2 \]

5. **Identify the Transition**:
Since the electron is transitioning to the \( n = 2 \) orbital, we need to identify the series:
- **Lyman series**: Transitions to \( n = 1 \)
- **Balmer series**: Transitions to \( n = 2 \)
- **Paschen series**: Transitions to \( n = 3 \)
- **Bracket series**: Transitions to \( n = 4 \)

Since the electron is ending in the \( n = 2 \) orbital, it must be part of the Balmer series.

### Final Answer:
The transition is associated with the **Balmer series**.

---