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 step by step, we will break it down into two parts: (a) finding the smallest wavelength emitted by hydrogen, and (b) listing the wavelengths emitted by hydrogen in the visible range.

### Part (a): Finding the Smallest Wavelength Emitted by Hydrogen

**Step 1: Identify the Transition**
- The problem states that the principal quantum number changes by 2. Starting from the ground state (n1 = 1), the next allowed state will be n2 = 3.

**Step 2: Calculate the Energy Difference**
- The energy of the transition can be calculated using the formula:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Substituting n1 = 1 and n2 = 3:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \, \text{eV} \left( 1 - \frac{1}{9} \right) = 13.6 \, \text{eV} \left( \frac{8}{9} \right) \]

**Step 3: Calculate the Energy Value**
- Calculate the energy:
\[ E = 13.6 \times \frac{8}{9} = \frac{108.8}{9} \approx 12.09 \, \text{eV} \]

**Step 4: Relate Energy to Wavelength**
- Use the equation \(E = \frac{hc}{\lambda}\) to find the wavelength:
\[ \lambda = \frac{hc}{E} \]
- Using \(h = 4.14 \times 10^{-15} \, \text{eV s}\) and \(c = 3 \times 10^8 \, \text{m/s}\):
\[ \lambda = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{12.09 \, \text{eV}} \]

**Step 5: Calculate the Wavelength**
- Calculate:
\[ \lambda \approx \frac{1.242 \times 10^{-6} \, \text{m}}{12.09} \approx 1.03 \times 10^{-7} \, \text{m} = 103 \, \text{nm} \]

### Part (b): List the Wavelengths Emitted by Hydrogen in the Visible Range

**Step 1: Identify the Transitions in the Visible Range**
- The visible range is from 380 nm to 780 nm. We will consider transitions starting from n1 = 2 (the first excited state).

**Step 2: Calculate the Energy for n1 = 2 and n2 = 4**
- For n1 = 2 and n2 = 4:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \, \text{eV} \left( \frac{4 - 1}{16} \right) = 13.6 \, \text{eV} \left( \frac{3}{16} \right) \]

**Step 3: Calculate the Energy Value**
- Calculate the energy:
\[ E = 13.6 \times \frac{3}{16} = \frac{40.8}{16} \approx 2.55 \, \text{eV} \]

**Step 4: Calculate the Wavelength**
- Using \(E = \frac{hc}{\lambda}\):
\[ \lambda = \frac{hc}{E} = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{2.55 \, \text{eV}} \approx \frac{1.242 \times 10^{-6} \, \text{m}}{2.55} \approx 487 \, \text{nm} \]

**Step 5: List Other Possible Transitions**
- Other transitions can be calculated similarly for n1 = 3 to n2 = 5, and so on, checking if they fall within the visible range.

### Summary of Wavelengths
- The smallest wavelength emitted by hydrogen is **103 nm**.
- The wavelength emitted in the visible range (for n1 = 2 to n2 = 4) is **487 nm**.