The wavelength of second Balmer line in hydrogen spectrum is 600nm.The wavelength for its 3rd line in Lyman series ? 1)800nm 2)120nm 3)400nm 4)200nm Answer is 2) 120nm Please explain Friends?

Question

Answer

First find out the value of 'R' using balmer series wavelength then proceed with Lyman series ￼

Answer

Answer

Understanding the Balmer and Lyman Series
The Balmer and Lyman series are sets of spectral lines resulting from electronic transitions in hydrogen. The Balmer series involves transitions to the second energy level (n=2), while the Lyman series involves transitions to the first energy level (n=1).
Given Information
- The wavelength of the second Balmer line (H-alpha) is 600 nm.
- We need to find the wavelength for the third line in the Lyman series (transition from n=3 to n=1).
Using the Rydberg Formula
The Rydberg formula can be used to calculate the wavelengths of spectral lines for hydrogen:
- For the Lyman series: 1/λ = R * (1 - 1/n²) where n = 2, 3, 4...
For the third line in the Lyman series (n=3 to n=1):
- n1 = 1 (lower level)
- n2 = 3 (upper level)
Applying the Rydberg Formula
1. Calculate using Rydberg constant (R = 1.097 × 10^7 m^-1):
1/λ = R * (1 - 1/3²)
1/λ = R * (1 - 1/9)
1/λ = R * (8/9)
2. Substitute R:
1/λ = (1.097 × 10^7) * (8/9)
3. Solve for λ:
λ = 9/(8 * 1.097 × 10^7)
λ ≈ 120 nm
Conclusion
Thus, the wavelength for the third line in the Lyman series is approximately 120 nm, confirming option 2.

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