A certain species of ionized atoms produces emission line spectral accorduing to the Bohr's model. A group of lines in the spectrum is forming a series in which in the shortest wavelength is `22.79 nm` and the longest wavelength is `41.02 nm`. The atomic number of atom is `Z`.
Based on above information, answer the following question:
The next to longest wavelength in the series of lines is

Options

A. `35.62 nm`
B. `30.47 nm`
C. `25.68 nm`
D. `12.64 nm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have the shortest wavelength (λ_min) = 22.79 nm and the longest wavelength (λ_max) = 41.02 nm. We need to find the next longest wavelength in the series of lines produced by the ionized atom.

### Step 2: Relate the wavelengths to energy transitions
According to the Bohr model, the energy of the emitted photon is inversely proportional to the wavelength. The longest wavelength corresponds to the smallest energy transition, which occurs between adjacent energy levels (n to n+1).

### Step 3: Set up the equations for λ_min and λ_max
1. For the longest wavelength (λ_max), the transition occurs from n to n+1:
\[ \frac{hc}{\lambda_{max}} = 13.6 Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \]

2. For the shortest wavelength (λ_min), the transition occurs from n to infinity:
\[ \frac{hc}{\lambda_{min}} = 13.6 Z^2 \left( \frac{1}{n^2} - 0 \right) \]

### Step 4: Divide the two equations
By dividing the two equations, we can eliminate constants:
\[ \frac{\lambda_{min}}{\lambda_{max}} = \frac{1 - \frac{n^2}{(n+1)^2}}{1} \]

### Step 5: Substitute the values of λ_min and λ_max
Substituting the values:
\[ \frac{22.79}{41.02} = 1 - \frac{n^2}{(n+1)^2} \]

### Step 6: Solve for n
1. Calculate the left side:
\[ \frac{22.79}{41.02} \approx 0.555 \]

2. Rearranging gives:
\[ 1 - 0.555 = \frac{n^2}{(n+1)^2} \]
\[ 0.445 = \frac{n^2}{(n+1)^2} \]

3. Cross-multiplying and simplifying:
\[ 0.445(n+1)^2 = n^2 \]
\[ 0.445(n^2 + 2n + 1) = n^2 \]
\[ 0.445n^2 + 0.89n + 0.445 = n^2 \]
\[ 0 = 0.555n^2 - 0.89n - 0.445 \]

4. Solving this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 0.555, b = -0.89, c = -0.445 \).

### Step 7: Calculate Z
Once we find n (let's say n = 2), we substitute it back to find Z:
\[ \frac{hc}{\lambda_{max}} = 13.6 Z^2 \left( \frac{1}{n^2} - 0 \right) \]
Solving this gives us Z = 4.

### Step 8: Find the next longest wavelength
The next longest wavelength corresponds to the transition from n=2 to n=3:
\[ \frac{hc}{\lambda} = 13.6 Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting Z = 4:
\[ \frac{hc}{\lambda} = 13.6 \times 16 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating this will yield the next longest wavelength.

### Final Answer
The next longest wavelength is approximately **30.47 nm**.

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