A sample of H atoms in certain excited state emits 10 different wavelengths in de-excitation to ground state. The maximum wavelength out of these wavelengths is:

Options

A. `4.05 xx10^(-6)m`
B. `4.05 xx10^(-4)m`
C. `4.5 xx10^(-6)m`
D. `4.5 xx10^(-4)m`

Answer

To solve the problem of finding the maximum wavelength emitted by a hydrogen atom transitioning from an excited state to the ground state, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A hydrogen atom in an excited state emits 10 different wavelengths as it transitions to the ground state. We need to find the maximum wavelength among these emitted wavelengths.

2. **Using the Formula for Spectral Lines**:
- The number of different spectral lines (or wavelengths) emitted when an electron transitions from an excited state (n) to the ground state (n=1) is given by the formula:
\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]
- Given that there are 10 different wavelengths, we can set up the equation:
\[ \frac{n(n-1)}{2} = 10 \]

3. **Solving for n**:
- Rearranging the equation gives:
\[ n(n-1) = 20 \]
- Testing integer values, we find:
- For \( n = 5 \):
\[ 5(5-1) = 5 \times 4 = 20 \]
- Thus, \( n = 5 \) is the excited state from which the electron transitions.

4. **Identifying the Transition for Maximum Wavelength**:
- The maximum wavelength corresponds to the smallest energy transition. In the case of hydrogen, this occurs when the electron transitions from \( n = 5 \) to \( n = 4 \).

5. **Using the Rydberg Formula**:
- The wavelength of the emitted light can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n_1 = 4 \), and \( n_2 = 5 \).

6. **Calculating the Wavelength**:
- Plugging in the values:
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]
- This simplifies to:
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{16} - \frac{1}{25} \right) \]
- Finding a common denominator (which is 80):
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{25 - 16}{400} \right) = R \left( \frac{9}{400} \right) \]
- Therefore:
\[ \lambda_{\text{max}} = \frac{400}{9R} \]

7. **Substituting the Rydberg Constant**:
- Substituting \( R = 1.097 \times 10^7 \, \text{m}^{-1} \):
\[ \lambda_{\text{max}} = \frac{400}{9 \times 1.097 \times 10^7} \]
- Calculating this gives:
\[ \lambda_{\text{max}} \approx 4.05 \times 10^{-6} \, \text{m} \]

### Final Answer:
The maximum wavelength emitted during the transition is approximately:
\[ \lambda_{\text{max}} \approx 4.05 \times 10^{-6} \, \text{m} \text{ or } 4050 \, \text{nm} \]