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The second member of Lyman's series is h...

The second member of Lyman's series is hydrogen spectrum has a wavelength of 5400 Å. Calculate the wavelength of the first member.

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To solve the problem of finding the wavelength of the first member of the Lyman series in the hydrogen spectrum, we can follow these steps: ### Step 1: Understand the Lyman Series The Lyman series corresponds to electronic transitions in a hydrogen atom where the electron falls to the n=1 energy level from higher energy levels (n=2, 3, 4,...). The first member of the Lyman series 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 electronic transitions is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \] where: - \( \lambda \) is the wavelength, - \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)), - \( n_f \) is the final energy level, - \( n_i \) is the initial energy level. ### Step 3: Identify the Values For the first member of the Lyman series: - \( n_f = 1 \) (final level), - \( n_i = 2 \) (initial level). ### Step 4: Substitute the Values into the Rydberg Formula Substituting the values into the Rydberg formula: \[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \] Calculating the right-hand side: \[ \frac{1}{\lambda_1} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \] ### Step 5: Calculate the Wavelength Now substituting the value of \( R \): \[ \frac{1}{\lambda_1} = 1.097 \times 10^7 \times \frac{3}{4} \] Calculating this gives: \[ \frac{1}{\lambda_1} = \frac{3.291 \times 10^7}{4} = 8.2275 \times 10^6 \, \text{m}^{-1} \] Now, taking the reciprocal to find \( \lambda_1 \): \[ \lambda_1 = \frac{1}{8.2275 \times 10^6} \approx 1.215 \times 10^{-7} \, \text{m} \] ### Step 6: Convert to Angstroms To convert meters to angstroms (1 Å = \( 10^{-10} \) m): \[ \lambda_1 \approx 1.215 \times 10^{-7} \, \text{m} = 1215 \, \text{Å} \] ### Final Answer The wavelength of the first member of the Lyman series is approximately **1215 Å**. ---
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