If `n_(1)` and `n_(2)` are the boundary value principal quantum numbers of a portion fo spectrum of emission spectrum of `H` atom, determine the wavelength ( in metre ) corresponding to last line ( longest `lambda)` . Given `:n_(1) + n_(2) =7, n_(2)-n_(1)=3` and `R_(H)=1.097xx10^(7)m^(-1)` . ( Give your answer in multiple of `10^(-6))`

Options

A. 0.8
B. 0.4
C. 0.1
D. 0.15

Explanation

To solve the problem, we need to determine the wavelength corresponding to the last line of the emission spectrum of the hydrogen atom given the conditions on the principal quantum numbers \( n_1 \) and \( n_2 \).

### Step-by-Step Solution:

1. **Identify the equations**: We are given two equations based on the principal quantum numbers:
\[ n_1 + n_2 = 7 \quad \text{(1)} \]
\[ n_2 - n_1 = 3 \quad \text{(2)} \]

2. **Solve for \( n_1 \) and \( n_2 \)**:
- From equation (1), we can express \( n_2 \) in terms of \( n_1 \):
\[ n_2 = 7 - n_1 \quad \text{(3)} \]
- Substitute equation (3) into equation (2):
\[ (7 - n_1) - n_1 = 3 \]
\[ 7 - 2n_1 = 3 \]
\[ 2n_1 = 4 \implies n_1 = 2 \]
- Now substitute \( n_1 \) back into equation (3) to find \( n_2 \):
\[ n_2 = 7 - 2 = 5 \]

3. **Use the Rydberg formula**: The Rydberg formula for the wavelength (\( \lambda \)) of the emitted light is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \).

4. **Substitute the values of \( n_1 \) and \( n_2 \)**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{25} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{25 - 4}{100} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{21}{100} \right) \]
\[ = 1.097 \times 10^7 \times 0.21 \]
\[ = 2.3037 \times 10^6 \, \text{m}^{-1} \]

5. **Calculate \( \lambda \)**:
\[ \lambda = \frac{1}{\frac{1}{\lambda}} = \frac{1}{2.3037 \times 10^6} \]
\[ \lambda \approx 4.344 \times 10^{-7} \, \text{m} \]

6. **Convert to the required format**:
\[ \lambda \approx 0.4344 \times 10^{-6} \, \text{m} \]
Rounding to one decimal place gives:
\[ \lambda \approx 0.4 \times 10^{-6} \, \text{m} \]

### Final Answer:
Thus, the wavelength corresponding to the last line of the emission spectrum of the hydrogen atom is:
\[ \lambda \approx 0.4 \times 10^{-6} \, \text{m} \]