Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelegnth equal to `18752 Å` is
(Rydberg constant `R = 1.097 xx 10^(7)` per metre)

Options

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

Explanation

To solve the problem, we need to determine which spectral series of the hydrogen spectrum corresponds to the given upper limit of wavelength, which is 18752 Å (or 18752 x 10^-10 m). We will use the Rydberg formula for hydrogen spectral lines.

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
\[ \lambda = 18752 \, \text{Å} = 18752 \times 10^{-10} \, \text{m} \]

2. **Use the Rydberg Formula**:
The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant \( (1.097 \times 10^7 \, \text{m}^{-1}) \), \( n_1 \) is the lower energy level, and \( n_2 \) is the upper energy level.

3. **Calculate \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} = \frac{1}{18752 \times 10^{-10}} \approx 5.33 \times 10^6 \, \text{m}^{-1} \]

4. **Substitute into the Rydberg Formula**:
\[ 5.33 \times 10^6 = 1.097 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

5. **Rearranging the Equation**:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{5.33 \times 10^6}{1.097 \times 10^7} \]
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \approx 0.486 \]

6. **Finding Suitable \( n_1 \) and \( n_2 \)**:
We know that for different series:
- Lyman series: \( n_1 = 1 \)
- Balmer series: \( n_1 = 2 \)
- Paschen series: \( n_1 = 3 \)
- Brackett series: \( n_1 = 4 \)

We will check combinations of \( n_1 \) and \( n_2 \) to satisfy the equation.

7. **Testing Values**:
- For \( n_1 = 3 \) and \( n_2 = 4 \):
\[ \frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \approx 0.0486 \]
This matches our earlier calculation.

8. **Conclusion**:
Since \( n_1 = 3 \) and \( n_2 = 4 \) corresponds to the Paschen series, we conclude that the spectral series with the upper limit of wavelength equal to 18752 Å is the **Paschen series**.

### Final Answer:
The spectral series which has an upper limit of wavelength equal to 18752 Å is the **Paschen series**.