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What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series?
- a)4 : 1
- b)4 : 3
- c)4 : 9
- d)5 : 9
Correct answer is option 'A'. Can you explain this answer?
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Explanation:
The Balmer series and the Lyman series are two different series of spectral lines in the hydrogen atom spectrum. The shortest wavelength in the Balmer series corresponds to the transition from n=3 to n=2 energy levels, while the shortest wavelength in the Lyman series corresponds to the transition from n=2 to n=1 energy levels.
Calculating the ratio:
- The formula for calculating the wavelength of spectral lines in hydrogen atom is given by:
- 1/λ = R(1/n₁² - 1/n₂²)
- Here, R is the Rydberg constant and n₁, n₂ are the principal quantum numbers.
- For the shortest wavelength in the Balmer series, n₁ = 3 and n₂ = 2.
- So, 1/λBalmer = R(1/2² - 1/3²)
- For the shortest wavelength in the Lyman series, n₁ = 2 and n₂ = 1.
- So, 1/λLyman = R(1/1² - 1/2²)
- Calculating the ratio:
- 1/λBalmer / 1/λLyman = (1/2² - 1/3²) / (1/1² - 1/2²)
= (1/4 - 1/9) / (1 - 1/4)
= (5/36) / (3/4)
= 5/9
Therefore, the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is 4 : 1. Option 'A' is the correct answer.
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What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series?a)4 : 1b)4 : 3c)4 : 9d)5 : 9Correct answer is option 'A'. Can you explain this answer?
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