The wavelength will be minimum for which of the following electronic transition in an unielectron species?

Options

A. `n = 6 " to " n = 4 `
B. `n = 4 " to " n = 2 `
C. `n = 3 " to " n = 1`
D. `n = 2 " to " n = 1`

Answer

To determine which electronic transition in a unielectron species corresponds to the minimum wavelength, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy difference (ΔE) between two electronic states is inversely proportional to the wavelength (λ) of the emitted or absorbed radiation. This relationship can be expressed as:
\[ \Delta E = \frac{hc}{\lambda} \]
Where:
- \(h\) is Planck's constant,
- \(c\) is the speed of light,
- \(\lambda\) is the wavelength.

From this equation, we can deduce that:
\[ \lambda \propto \frac{1}{\Delta E} \]
This means that if ΔE is larger, λ will be smaller (and vice versa).

### Step 2: Calculate ΔE for each transition
For a unielectron species, the energy difference can be calculated using the formula:
\[ \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
Where \(n_f\) is the final energy level and \(n_i\) is the initial energy level.

### Step 3: Evaluate each option
Let’s evaluate the ΔE for each transition:

1. **Option A**: Transition from \(n_i = 6\) to \(n_f = 4\)
\[ \Delta E_A = 2.18 \times 10^{-18} \left( \frac{1}{4^2} - \frac{1}{6^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{16} - \frac{1}{36} \right) \]
\[ = 2.18 \times 10^{-18} \left( \frac{9 - 4}{144} \right) = 2.18 \times 10^{-18} \times \frac{5}{144} = 7.57 \times 10^{-20} \text{ J} \]

2. **Option B**: Transition from \(n_i = 4\) to \(n_f = 2\)
\[ \Delta E_B = 2.18 \times 10^{-18} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ = 2.18 \times 10^{-18} \left( \frac{4 - 1}{16} \right) = 2.18 \times 10^{-18} \times \frac{3}{16} = 4.09 \times 10^{-19} \text{ J} \]

3. **Option C**: Transition from \(n_i = 3\) to \(n_f = 1\)
\[ \Delta E_C = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 2.18 \times 10^{-18} \left( 1 - \frac{1}{9} \right) \]
\[ = 2.18 \times 10^{-18} \left( \frac{8}{9} \right) = 1.93 \times 10^{-18} \text{ J} \]

4. **Option D**: Transition from \(n_i = 2\) to \(n_f = 1\)
\[ \Delta E_D = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 2.18 \times 10^{-18} \left( 1 - \frac{1}{4} \right) \]
\[ = 2.18 \times 10^{-18} \left( \frac{3}{4} \right) = 1.64 \times 10^{-18} \text{ J} \]

### Step 4: Compare the energy differences
Now, we can compare the calculated ΔE values:
- ΔE_A = \(7.57 \times 10^{-20} \text{ J}\)
- ΔE_B = \(4.09 \times 10^{-19} \text{ J}\)
- ΔE_C = \(1.93 \times 10^{-18} \text{ J}\)
- ΔE_D = \(1.64 \times 10^{-18} \text{ J}\)

### Conclusion
The transition with the highest ΔE corresponds to the minimum wavelength. In this case, **Option C (from \(n=3\) to \(n=1\))** has the highest energy difference, thus it has the minimum wavelength.