Find the two longest wavelength `("in" Å)` emitted when hydrogen atom make transition and the spectrum lines lie in the visible region `(R = 1.097 xx 10^(7) m^(-1))`

Explanation

To find the two longest wavelengths emitted by a hydrogen atom when it makes transitions in the visible region, we will focus on the Balmer series. The Balmer series corresponds to transitions where the final energy level (n1) is 2, and the initial energy levels (n2) are greater than 2.

### Step-by-Step Solution:

1. **Identify the Series**:
The visible spectrum lines for hydrogen are part of the Balmer series, where the transitions end at n=2.

2. **Determine the Transitions**:
The longest wavelengths correspond to the smallest energy differences, which occur for the transitions from n=3 to n=2 and n=4 to n=2.

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

4. **Calculate the Longest Wavelength (n=3 to n=2)**:
- For the transition from n=3 to n=2:
\[ \frac{1}{\lambda_{3 \to 2}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{9 - 4}{36} \right) = 1.097 \times 10^7 \left( \frac{5}{36} \right) \]
\[ = 1.097 \times 10^7 \times 0.138888 \approx 6.5628 \times 10^{-7} \, \text{m} \]
- Convert to angstroms:
\[ \lambda_{3 \to 2} = 6.5628 \times 10^{-7} \, \text{m} = 6562.8 \, \text{Å} \]

5. **Calculate the Second Longest Wavelength (n=4 to n=2)**:
- For the transition from n=4 to n=2:
\[ \frac{1}{\lambda_{4 \to 2}} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{4 - 1}{16} \right) = 1.097 \times 10^7 \left( \frac{3}{16} \right) \]
\[ = 1.097 \times 10^7 \times 0.1875 \approx 4.8617 \times 10^{-7} \, \text{m} \]
- Convert to angstroms:
\[ \lambda_{4 \to 2} = 4.8617 \times 10^{-7} \, \text{m} = 4861.7 \, \text{Å} \]

6. **Final Answer**:
The two longest wavelengths emitted when a hydrogen atom makes transitions in the visible region are:
- From n=3 to n=2: **6562.8 Å**
- From n=4 to n=2: **4861.7 Å**