The shortest wavelength for Lyman series is 912 Å. What will be the longest wavelength in Paschen series?

Options

A. `1216 Å`
B. `3646 Å`
C. `18761 Å`
D. `8208 Å`

Answer

To find the longest wavelength in the Paschen series given that the shortest wavelength for the Lyman series is 912 Å, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions in a hydrogen atom where the electron falls to the n=1 energy level from higher energy levels (n=2, 3, 4, ...). The shortest wavelength occurs when the electron transitions from infinity (n=∞) to n=1.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of light emitted during these transitions is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

For the Lyman series, \( n_1 = 1 \) and \( n_2 = \infty \).

### Step 3: Calculate the Rydberg Constant
Given that the shortest wavelength for the Lyman series is 912 Å, we can set up the equation:

\[ \frac{1}{912} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]

Since \( \frac{1}{\infty^2} = 0 \), this simplifies to:

\[ \frac{1}{912} = R (1 - 0) \implies R = \frac{1}{912} \]

### Step 4: Understand the Paschen Series
The Paschen series corresponds to transitions where the electron falls to the n=3 energy level from higher levels (n=4, 5, 6, ...). The longest wavelength occurs when the electron transitions from n=4 to n=3.

### Step 5: Apply the Rydberg Formula for the Paschen Series
For the longest wavelength in the Paschen series:

\[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

Substituting the values:

\[ \frac{1}{\lambda} = \frac{1}{912} \left( \frac{1}{9} - \frac{1}{16} \right) \]

### Step 6: Calculate the Difference
Calculate \( \frac{1}{9} - \frac{1}{16} \):

\[ \frac{1}{9} = \frac{16}{144}, \quad \frac{1}{16} = \frac{9}{144} \]

Thus,

\[ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \]

### Step 7: Substitute Back to Find λ
Now substitute this back into the equation:

\[ \frac{1}{\lambda} = \frac{1}{912} \cdot \frac{7}{144} \]

This simplifies to:

\[ \lambda = \frac{912 \cdot 144}{7} \]

### Step 8: Calculate the Final Value
Calculating this gives:

\[ \lambda = \frac{131328}{7} \approx 18761.14 \text{ Å} \]

### Final Answer
Thus, the longest wavelength in the Paschen series is approximately **18761 Å**.

---