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The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3
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The frequency of light emitted, for the transition n=4 to n=2 of He+ i...
Frequency of Light Emitted in He (n=4 to n=2) Transition
To determine the frequency of light emitted in the transition from n=4 to n=2 in He, we need to apply the Rydberg formula, which relates the wavelength (λ) or frequency (ν) of light emitted or absorbed during an electronic transition to the initial (n₁) and final (n₂) energy levels of the atom.
The Rydberg formula is given by:
1/λ = R * (1/n₁² - 1/n₂²)
where λ is the wavelength, R is the Rydberg constant, and n₁ and n₂ are the initial and final energy levels, respectively.
Comparing the Transition in He to H Atom
To determine the corresponding transition in the hydrogen (H) atom, we need to find a transition that has the same frequency as the n=4 to n=2 transition in He.
Calculating Frequencies for Each Transition
1) Transition: n=3 to n=12
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/3² - 1/12²)
1/λ = R * (1/9 - 1/144)
1/λ = R * (143/1296)
λ = 1296/143R
2) Transition: n=2 to n=1
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/2² - 1/1²)
1/λ = R * (1/4 - 1/1)
1/λ = R * (3/4)
λ = 4/3R
3) Transition: n=3 to n=1
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/3² - 1/1²)
1/λ = R * (1/9 - 1/1)
1/λ = R * (8/9)
λ = 9/8R
4) Transition: n=4 to n=3
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/4² - 1/3²)
1/λ = R * (1/16 - 1/9)
1/λ = R * (9/144 - 16/144)
1/λ = R * (-7/144)
λ = -144/7R
Comparison of Frequencies
Comparing the frequency of light emitted in the He (n=4 to n=2) transition
To determine the frequency of light emitted in the transition from n=4 to n=2 in He, we need to apply the Rydberg formula, which relates the wavelength (λ) or frequency (ν) of light emitted or absorbed during an electronic transition to the initial (n₁) and final (n₂) energy levels of the atom.
The Rydberg formula is given by:
1/λ = R * (1/n₁² - 1/n₂²)
where λ is the wavelength, R is the Rydberg constant, and n₁ and n₂ are the initial and final energy levels, respectively.
Comparing the Transition in He to H Atom
To determine the corresponding transition in the hydrogen (H) atom, we need to find a transition that has the same frequency as the n=4 to n=2 transition in He.
Calculating Frequencies for Each Transition
1) Transition: n=3 to n=12
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/3² - 1/12²)
1/λ = R * (1/9 - 1/144)
1/λ = R * (143/1296)
λ = 1296/143R
2) Transition: n=2 to n=1
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/2² - 1/1²)
1/λ = R * (1/4 - 1/1)
1/λ = R * (3/4)
λ = 4/3R
3) Transition: n=3 to n=1
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/3² - 1/1²)
1/λ = R * (1/9 - 1/1)
1/λ = R * (8/9)
λ = 9/8R
4) Transition: n=4 to n=3
Using the Rydberg formula, we can calculate the frequency of light emitted in this transition in the hydrogen atom:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = R * (1/4² - 1/3²)
1/λ = R * (1/16 - 1/9)
1/λ = R * (9/144 - 16/144)
1/λ = R * (-7/144)
λ = -144/7R
Comparison of Frequencies
Comparing the frequency of light emitted in the He (n=4 to n=2) transition
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The frequency of light emitted, for the transition n=4 to n=2 of He+ i...
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The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3
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The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3 for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3 covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3.
The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3 for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3 covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3.
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The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3, a detailed solution for The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3 has been provided alongside types of The frequency of light emitted, for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following:1) n=3 to n=12) n=2 to n=13) n=3 to n=14) n=4 to n=3 theory, EduRev gives you an
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