The limiting line Balmer series will have a frequency of

Options

A. `32.29 xx 10^(15) s^(-1)`
B. `3.65 xx 10^(15) s^(-1)`
C. `-8.22 xx 10^(15) s^(-1)`
D. `8.22 xx 10^(15) s^(-1)`

Explanation

To find the limiting line frequency of the Balmer series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Balmer Series**: The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The limiting line occurs when n2 approaches 1 (the lowest energy level) and n1 approaches infinity.

2. **Use the Rydberg Formula**: The frequency (ν) of the emitted light can be calculated using the Rydberg formula:
\[ \nu = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant for hydrogen, approximately \( 3.29 \times 10^{15} \, \text{s}^{-1} \).
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)).
- \( n_1 \) is the lower energy level (for the Balmer series, \( n_1 = 2 \)).
- \( n_2 \) is the higher energy level (for the limiting line, \( n_2 \to \infty \)).

3. **Substitute the Values**: In the case of the limiting line of the Balmer series:
- \( n_1 = 2 \) (which gives \( \frac{1}{n_1^2} = \frac{1}{4} \))
- \( n_2 \to \infty \) (which gives \( \frac{1}{n_2^2} \to 0 \))

Therefore, the formula simplifies to:
\[ \nu = R_H \cdot \left( \frac{1}{4} - 0 \right) = R_H \cdot \frac{1}{4} \]

4. **Calculate the Frequency**: Now substituting the value of \( R_H \):
\[ \nu = 3.29 \times 10^{15} \cdot \frac{1}{4} = 8.225 \times 10^{14} \, \text{s}^{-1} \]

5. **Final Result**: Rounding this value gives us:
\[ \nu \approx 8.22 \times 10^{14} \, \text{s}^{-1} \]

Thus, the limiting line of the Balmer series will have a frequency of approximately \( 8.22 \times 10^{14} \, \text{s}^{-1} \).