The hydrogen atom in the ground state is excited by means of monochromatic radiation of wavelength `xA_(0)` The resulting spectrum consists of 15 different lines. Calculate the value of x.

Explanation

To solve the problem, we need to determine the wavelength \( x \) of monochromatic radiation that excites a hydrogen atom from its ground state, resulting in 15 different spectral lines.

### Step-by-Step Solution:

1. **Understanding the Number of Lines**:
The number of spectral lines \( N \) produced when an electron transitions from a higher energy level \( n_2 \) to a lower energy level \( n_1 \) is given by the formula:
\[ N = \frac{n_2(n_2 - n_1)}{2} \]
Here, since the electron is excited from the ground state (where \( n_1 = 1 \)), we can set \( n_1 = 1 \).

2. **Setting Up the Equation**:
Given that \( N = 15 \) and \( n_1 = 1 \), we can substitute these values into the equation:
\[ 15 = \frac{n_2(n_2 - 1)}{2} \]
Multiplying both sides by 2 gives:
\[ 30 = n_2(n_2 - 1) \]
Rearranging this, we get:
\[ n_2^2 - n_2 - 30 = 0 \]

3. **Solving the Quadratic Equation**:
We can solve the quadratic equation \( n_2^2 - n_2 - 30 = 0 \) using the quadratic formula:
\[ n_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = -30 \):
\[ n_2 = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \]
\[ n_2 = \frac{1 \pm \sqrt{1 + 120}}{2} \]
\[ n_2 = \frac{1 \pm \sqrt{121}}{2} \]
\[ n_2 = \frac{1 \pm 11}{2} \]
This gives us two possible solutions:
\[ n_2 = 6 \quad \text{(since n must be positive)} \]

4. **Calculating the Wavelength**:
Now that we have \( n_2 = 6 \) and \( n_1 = 1 \), we can use the Rydberg formula to calculate the wavelength \( \lambda \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For hydrogen, \( R = 1.097 \times 10^7 \, \text{m}^{-1} \). Substituting the values:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{6^2} \right) \]
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{36} \right) \]
\[ \frac{1}{\lambda} = R \left( \frac{36 - 1}{36} \right) = R \left( \frac{35}{36} \right) \]
\[ \frac{1}{\lambda} = \frac{35}{36} R \]

5. **Finding the Wavelength**:
Now, substituting the value of \( R \):
\[ \lambda = \frac{36}{35 R} \]
Converting \( R \) into angstroms:
\[ R = 1.097 \times 10^7 \, \text{m}^{-1} = 912 \, \text{Å}^{-1} \]
Thus,
\[ \lambda = \frac{36}{35 \times 912} \, \text{Å} \]
\[ \lambda = \frac{36}{31920} \approx 0.00113 \, \text{Å}^{-1} \]
Finally, calculating gives us:
\[ \lambda \approx 2937.3 \, \text{Å} \]

### Final Answer:
The value of \( x \) is approximately \( 2937.3 \, \text{Å} \).