A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelength of incident radiation should be

Options

A. 800 A
B. 825 A
C. 970 A
D. 1025 A

Answer

To solve the problem of finding the wavelength of incident radiation required to get six lines in the emission spectrum of a hydrogen atom in the ground state, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Emission Spectrum**:
The number of spectral lines emitted when an electron transitions from a higher energy level (n) to the ground state (n=1) can be calculated using the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]

2. **Setting Up the Equation**:
Given that we want 6 spectral lines, we can set up the equation:
\[ 6 = \frac{n(n-1)}{2} \]

3. **Solving for n**:
To find n, we can rearrange the equation:
\[ 12 = n(n-1) \]
This simplifies to:
\[ n^2 - n - 12 = 0 \]
We can factor this quadratic equation:
\[ (n - 4)(n + 3) = 0 \]
Thus, \( n = 4 \) (since n cannot be negative).

4. **Finding the Wavelength**:
Now, we need to find the wavelength of the radiation corresponding to the transition from n=4 to n=1. We use the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n_1 = 1 \), and \( n_2 = 4 \).

5. **Substituting Values**:
Substituting the values into the formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{16 - 1}{16} \right) = R \left( \frac{15}{16} \right) \]

6. **Calculating λ**:
Thus,
\[ \frac{1}{\lambda} = R \cdot \frac{15}{16} \]
Therefore,
\[ \lambda = \frac{16}{15R} \]

7. **Substituting Rydberg Constant**:
Now, substituting \( R \):
\[ \lambda = \frac{16}{15 \cdot 1.097 \times 10^7} \approx 970 \, \text{Å} \]

### Final Answer:
The wavelength of the incident radiation required to get six lines in the emission spectrum of a hydrogen atom in the ground state is approximately **970 Å**.

---