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Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.
- a)4
- b)3
- c)5
- d)2
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Electron emits radiation with a wavelength of 4900 x10-10 m from a cer...
Given:
According to the question,
We know that,
Since the radiation is in the Balmer series (Visible region)
Therefore, n1=2
Hence, this is the correct option.
According to the question,
We know that,
Since the radiation is in the Balmer series (Visible region)
Therefore, n1=2
Hence, this is the correct option.
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Community Answer
Electron emits radiation with a wavelength of 4900 x10-10 m from a cer...
Given Data:
- Wavelength of radiation emitted by electron: 4900 x 10^-10 m
- RH (Rydberg constant): 1.097 x 10^7 m^-1
Formula Used:
The formula to calculate the wavelength of radiation emitted in the Balmer series is given by:
1/λ = RH (1/n1^2 - 1/n2^2)
Solving the equation:
- Given wavelength = 4900 x 10^-10 m
- RH = 1.097 x 10^7 m^-1
- Let n1 = 2 (for Balmer series)
- Let n2 = n (to be determined)
Plugging in the values in the formula:
1/(4900 x 10^-10) = 1.097 x 10^7 (1/2^2 - 1/n^2)
Solving further gives:
n = 4
Therefore, the electronic state of the electron emitting radiation with a wavelength of 4900 x 10^-10 m in the Balmer series is 4. Hence, the correct answer is option 'A'.
- Wavelength of radiation emitted by electron: 4900 x 10^-10 m
- RH (Rydberg constant): 1.097 x 10^7 m^-1
Formula Used:
The formula to calculate the wavelength of radiation emitted in the Balmer series is given by:
1/λ = RH (1/n1^2 - 1/n2^2)
Solving the equation:
- Given wavelength = 4900 x 10^-10 m
- RH = 1.097 x 10^7 m^-1
- Let n1 = 2 (for Balmer series)
- Let n2 = n (to be determined)
Plugging in the values in the formula:
1/(4900 x 10^-10) = 1.097 x 10^7 (1/2^2 - 1/n^2)
Solving further gives:
n = 4
Therefore, the electronic state of the electron emitting radiation with a wavelength of 4900 x 10^-10 m in the Balmer series is 4. Hence, the correct answer is option 'A'.
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Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer?
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Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. 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 Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer?.
Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. 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 Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer?.
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Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Electron emits radiation with a wavelength of 4900 x10-10 m from a certain electronic state and is lying in the Balmer series in the spectrum of hydrogen atom. Find the electronic state of the electron. Take approximate value and RH = 1.097 x 107m-1.a)4b)3c)5d)2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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