The hydrogen atom in the ground state is excited by mass of monochromatic radiations of wavelength
` lambda Å ` . The resulting spectrum consists of maximum `15` different lines . What is the value of ` lambda` ? (`R_H = 109737 cm^(-1))`.

Options

A. `937.3 Å`
B. `1025 Å`
C. `1236 Å`
D. None of these

Answer

To solve the problem, we need to determine the wavelength of monochromatic radiation that excites a hydrogen atom from the ground state to a higher energy level, resulting in a spectrum with 15 different lines.

### Step-by-Step Solution:

1. **Understanding the Number of Spectral Lines**:
The number of spectral lines (N) produced when an electron transitions between energy levels can be calculated using the formula:
\[ N = \frac{n(n-1)}{2} \]
where \( n \) is the number of energy levels involved in the transitions.

2. **Setting Up the Equation**:
Given that \( N = 15 \), we can set up the equation:
\[ 15 = \frac{n(n-1)}{2} \]
Multiplying both sides by 2 gives:
\[ 30 = n(n-1) \]

3. **Finding n**:
We need to find integer values of \( n \) that satisfy the equation \( n(n-1) = 30 \). Testing values:
- For \( n = 6 \):
\[ 6(6-1) = 6 \times 5 = 30 \]
Thus, \( n = 6 \).

4. **Identifying Energy Levels**:
The electron transitions from the ground state (\( n_1 = 1 \)) to the sixth energy level (\( n_2 = 6 \)).

5. **Using the Rydberg Formula**:
The wavelength (\( \lambda \)) of the emitted or absorbed radiation can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H = 109737 \, \text{cm}^{-1} \).

6. **Substituting Values**:
Substituting \( n_1 = 1 \) and \( n_2 = 6 \) into the formula:
\[ \frac{1}{\lambda} = 109737 \left( \frac{1}{1^2} - \frac{1}{6^2} \right) \]
\[ = 109737 \left( 1 - \frac{1}{36} \right) \]
\[ = 109737 \left( \frac{36 - 1}{36} \right) \]
\[ = 109737 \left( \frac{35}{36} \right) \]

7. **Calculating \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} = 109737 \times \frac{35}{36} \]
\[ = 109737 \times 0.9722 \approx 106703.5 \, \text{cm}^{-1} \]

8. **Finding \( \lambda \)**:
\[ \lambda = \frac{1}{106703.5} \, \text{cm} \approx 9.373 \times 10^{-6} \, \text{cm} \]

9. **Converting to Angstroms**:
Since \( 1 \, \text{cm} = 10^{10} \, \text{Å} \):
\[ \lambda \approx 9.373 \times 10^{-6} \, \text{cm} \times 10^{10} \, \text{Å/cm} = 9373 \, \text{Å} \]

10. **Final Answer**:
The wavelength of the monochromatic radiation is approximately:
\[ \lambda \approx 937.3 \, \text{Å} \]