Calculate the shortest wavelength in Paschen series if the longest wavelength in Balmar series is 6563 `A@`.

Answer

To calculate the shortest wavelength in the Paschen series given that the longest wavelength in the Balmer series is 6563 Å, we will follow these steps:

### Step 1: Understand the series and their transitions
The Balmer series corresponds to transitions where the final energy level (n) is 2, and the transitions start from higher energy levels (m). The longest wavelength in the Balmer series occurs when the transition is from n=3 to n=2.

### Step 2: Use the formula for wavelength in the Balmer series
The formula for the wavelength (λ) in the Balmer series is given by:

\[ \frac{1}{\lambda_B} = R \cdot (1 - \frac{1}{n^2}) \]

Where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n \) is the principal quantum number of the initial state.

For the longest wavelength in the Balmer series (λ_B = 6563 Å), we have:

\[ \frac{1}{\lambda_B} = R \cdot (1 - \frac{1}{3^2}) = R \cdot (1 - \frac{1}{9}) = R \cdot \frac{8}{9} \]

### Step 3: Calculate the wavelength in the Paschen series
The Paschen series corresponds to transitions where the final energy level (n) is 3. The shortest wavelength in the Paschen series occurs when the transition starts from m = ∞ (infinity) to n = 3.

Using the formula for the Paschen series:

\[ \frac{1}{\lambda_P} = R \cdot (1 - \frac{1}{3^2}) = R \cdot (1 - \frac{1}{9}) = R \cdot \frac{8}{9} \]

### Step 4: Relate the wavelengths of the two series
From the previous steps, we have:

\[ \frac{1}{\lambda_B} = R \cdot \frac{8}{9} \]
\[ \frac{1}{\lambda_P} = R \cdot \frac{8}{9} \]

To find the relationship between the two wavelengths, we can express λ_P in terms of λ_B:

\[ \lambda_P = \frac{9}{8} \cdot \lambda_B \]

### Step 5: Substitute the value of λ_B
Now, substituting the value of λ_B (6563 Å):

\[ \lambda_P = \frac{9}{8} \cdot 6563 \, \text{Å} \]

### Step 6: Calculate λ_P
Calculating the above expression:

\[ \lambda_P = \frac{9 \cdot 6563}{8} = \frac{59067}{8} = 7383.375 \, \text{Å} \]

Thus, the shortest wavelength in the Paschen series is approximately:

\[ \lambda_P \approx 7383.38 \, \text{Å} \]

### Final Answer
The shortest wavelength in the Paschen series is approximately **7383.38 Å**.

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