The ratio of te shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are:

Options

A. Balmer and Barckett
B. Paschen and Pfund
C. Brackett and Pfund
D. Lyman and Paschen

Explanation

To solve the problem, we need to determine the two spectral series of the hydrogen spectrum whose shortest wavelength ratio is approximately 9. We will use the concept of energy levels in the hydrogen atom and the relationship between energy and wavelength.

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
The energy of an electron in a hydrogen atom at a principal quantum number \( n \) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Finding the Shortest Wavelength**:
The shortest wavelength (or maximum energy transition) in a spectral series occurs when an electron transitions from the \( n = 1 \) level to \( n = \infty \). The energy change for this transition is:
\[ \Delta E = E_{\infty} - E_n = 0 - \left(-\frac{13.6}{n^2}\right) = \frac{13.6}{n^2} \, \text{eV} \]

3. **Relating Energy and Wavelength**:
The energy of a photon is related to its wavelength \( \lambda \) by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. Thus, we can write:
\[ \lambda = \frac{hc}{E} \]

4. **Calculating the Shortest Wavelength for Two Series**:
Let's denote the two series as \( n_1 \) and \( n_2 \). The shortest wavelengths for these series can be expressed as:
\[ \lambda_1 = \frac{hc}{\Delta E_1} = \frac{hc}{\frac{13.6}{n_1^2}} = \frac{hc \cdot n_1^2}{13.6} \]
\[ \lambda_2 = \frac{hc}{\Delta E_2} = \frac{hc}{\frac{13.6}{n_2^2}} = \frac{hc \cdot n_2^2}{13.6} \]

5. **Setting Up the Ratio**:
The problem states that the ratio of the shortest wavelengths is approximately 9:
\[ \frac{\lambda_1}{\lambda_2} = 9 \]
Substituting the expressions for \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{n_1^2}{n_2^2} = 9 \]
Taking the square root of both sides gives:
\[ \frac{n_1}{n_2} = \frac{1}{3} \]
Thus, we can express this as:
\[ n_2 = 3n_1 \]

6. **Identifying the Series**:
The possible values for \( n_1 \) and \( n_2 \) must correspond to known spectral series:
- For \( n_1 = 1 \) (Lyman series), \( n_2 = 3 \) (Paschen series).
- For \( n_1 = 2 \) (Balmer series), \( n_2 = 6 \) (not a valid series).
- For \( n_1 = 3 \) (Bracket series), \( n_2 = 9 \) (not a valid series).
- For \( n_1 = 4 \) (Pfund series), \( n_2 = 12 \) (not a valid series).

The only valid pair is \( n_1 = 1 \) and \( n_2 = 3 \), corresponding to the Lyman series and Paschen series.

### Final Answer:
The spectral series are **Lyman and Paschen**.