Suppose in certine condition only those transition are allowed to hydrogen atoms in which the principal quantum number a changes by`2` (a) Find the smaller wavelength emitted by hydrogen (b) list the wavelength emitted by hydrogen in the visible range `(380 nm to 780 nm)`

Explanation

To solve the problem, we will break it down into two parts as per the question.

### Part (a): Find the smaller wavelength emitted by hydrogen.

1. **Identify the allowed transitions**: The problem states that only transitions where the principal quantum number changes by 2 are allowed. Therefore, the possible transitions are:
- From n = 3 to n = 1 (Δn = 2)
- From n = 4 to n = 2 (Δn = 2)
- From n = 5 to n = 3 (Δn = 2)

2. **Use the Rydberg formula**: The wavelength (λ) of the emitted light can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) = Rydberg constant = \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( Z \) = atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy states, respectively.

3. **Calculate for the transition from n = 3 to n = 1**:
- Here, \( n_1 = 1 \) and \( n_2 = 3 \):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ = 1.097 \times 10^7 \left( 1 - \frac{1}{9} \right) = 1.097 \times 10^7 \left( \frac{8}{9} \right) \]
\[ = 1.097 \times 10^7 \times \frac{8}{9} = 9.75 \times 10^6 \, \text{m}^{-1} \]

4. **Calculate the wavelength**:
\[ \lambda = \frac{1}{9.75 \times 10^6} \approx 1.027 \times 10^{-7} \, \text{m} = 102.7 \, \text{nm} \]
Therefore, the smaller wavelength emitted by hydrogen is approximately **102.7 nm**.

### Part (b): List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm).

1. **Calculate for the transition from n = 4 to n = 2**:
- Here, \( n_1 = 2 \) and \( n_2 = 4 \):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \left( \frac{4 - 1}{16} \right) = 1.097 \times 10^7 \times \frac{3}{16} \]
\[ = 2.06 \times 10^6 \, \text{m}^{-1} \]
\[ \lambda = \frac{1}{2.06 \times 10^6} \approx 486.5 \, \text{nm} \]
This wavelength is in the visible range.

2. **Calculate for the transition from n = 5 to n = 3**:
- Here, \( n_1 = 3 \) and \( n_2 = 5 \):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{25} \right) = 1.097 \times 10^7 \left( \frac{25 - 9}{225} \right) = 1.097 \times 10^7 \times \frac{16}{225} \]
\[ = 7.79 \times 10^6 \, \text{m}^{-1} \]
\[ \lambda = \frac{1}{7.79 \times 10^6} \approx 128.3 \, \text{nm} \]
This wavelength is not in the visible range.

### Summary of Wavelengths in the Visible Range:
- The only wavelength emitted by hydrogen in the visible range is approximately **486.5 nm** (from the transition n = 4 to n = 2).