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What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series ? (h = Planck constant; c = Velocity of light; R = Rydberg constant).
- a)5hcr/36
- b)4hcr/3
- c)3hcr/4
- d)7hcr/144
Correct answer is option 'C'. Can you explain this answer?
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What is the lowest energy of the spectral line emitted by the hydrogen...
Calculation of Lowest Energy in Lyman Series
The Lyman series corresponds to the transition of electrons from higher energy levels to the n=1 energy level of hydrogen atom. The energy of the spectral line emitted in Lyman series can be calculated using the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength of the spectral line emitted, R is the Rydberg constant, and n1 and n2 are the initial and final energy levels, respectively.
For the Lyman series, n1 = 2 and n2 = 1. Therefore, the formula becomes:
1/λ = R(1/2^2 - 1/1^2)
Simplifying the formula, we get:
λ = h c R/3
where h is the Planck constant and c is the velocity of light.
The lowest energy in the Lyman series corresponds to the longest wavelength, which can be obtained by taking the limit as n2 approaches infinity. In this limit, we get:
1/λ = R(1/2^2 - 0)
λ = h c R/4
Therefore, the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series is given by:
E = hc/λ = 4hcr/3
Hence, the correct option is C.
The Lyman series corresponds to the transition of electrons from higher energy levels to the n=1 energy level of hydrogen atom. The energy of the spectral line emitted in Lyman series can be calculated using the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength of the spectral line emitted, R is the Rydberg constant, and n1 and n2 are the initial and final energy levels, respectively.
For the Lyman series, n1 = 2 and n2 = 1. Therefore, the formula becomes:
1/λ = R(1/2^2 - 1/1^2)
Simplifying the formula, we get:
λ = h c R/3
where h is the Planck constant and c is the velocity of light.
The lowest energy in the Lyman series corresponds to the longest wavelength, which can be obtained by taking the limit as n2 approaches infinity. In this limit, we get:
1/λ = R(1/2^2 - 0)
λ = h c R/4
Therefore, the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series is given by:
E = hc/λ = 4hcr/3
Hence, the correct option is C.
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What is the lowest energy of the spectral line emitted by the hydrogen...
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What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series ? (h = Planck constant; c = Velocity of light; R = Rydberg constant).a)5hcr/36b)4hcr/3c)3hcr/4d)7hcr/144Correct answer is option 'C'. Can you explain this answer?
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What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series ? (h = Planck constant; c = Velocity of light; R = Rydberg constant).a)5hcr/36b)4hcr/3c)3hcr/4d)7hcr/144Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series ? (h = Planck constant; c = Velocity of light; R = Rydberg constant).a)5hcr/36b)4hcr/3c)3hcr/4d)7hcr/144Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series ? (h = Planck constant; c = Velocity of light; R = Rydberg constant).a)5hcr/36b)4hcr/3c)3hcr/4d)7hcr/144Correct answer is option 'C'. Can you explain this answer?.
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