The shortest wavelength in Lyman series is 91.2 nm. The longest wavelength of the series is

Options

A. 121.6 nm
B. 182.4 nm
C. 234.4 nm
D. 364.8 nm

Answer

To find the longest wavelength in the Lyman series, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electronic transitions in hydrogen where an electron falls to the first energy level (n=1) from higher energy levels (n=2, n=3, n=4, ...). The shortest wavelength corresponds to the transition from n = ∞ to n = 1, while the longest wavelength corresponds to the transition from n = 2 to n = 1.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted light 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 (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level (for Lyman series, \( n_1 = 1 \)),
- \( n_2 \) is the higher energy level.

### Step 3: Identify the Wavelengths
1. **Shortest Wavelength**: This corresponds to the transition from \( n = \infty \) to \( n = 1 \):
\[ \frac{1}{\lambda_{\text{shortest}}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]
Given that the shortest wavelength \( \lambda_{\text{shortest}} = 91.2 \, \text{nm} \), we can express this as:
\[ \frac{1}{91.2 \times 10^{-9}} = R \]

2. **Longest Wavelength**: This corresponds to the transition from \( n = 2 \) to \( n = 1 \):
\[ \frac{1}{\lambda_{\text{longest}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

### Step 4: Calculate the Longest Wavelength
From the previous step, we can express the longest wavelength as:
\[ \frac{1}{\lambda_{\text{longest}}} = \frac{3R}{4} \]
Substituting \( R \) from the shortest wavelength:
\[ \frac{1}{\lambda_{\text{longest}}} = \frac{3}{4} \left( \frac{1}{91.2 \times 10^{-9}} \right) \]
Now, solve for \( \lambda_{\text{longest}} \):
\[ \lambda_{\text{longest}} = \frac{4 \times 91.2 \times 10^{-9}}{3} \]
Calculating this gives:
\[ \lambda_{\text{longest}} = \frac{364.8 \times 10^{-9}}{3} = 121.6 \times 10^{-9} \, \text{m} = 121.6 \, \text{nm} \]

### Final Answer
The longest wavelength in the Lyman series is **121.6 nm**.

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