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The shortest wavelength in Lyman series ...

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

A

121.6 nm

B

182.4 nm

C

234.4 nm

D

364.8 nm

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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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