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In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]
Planck’s constant, h = 6.63 × 10–34 Js
  • a)
    5 → 2
  • b)
    4 → 1
  • c)
    2 → 5
  • d)
    3 → 2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In Bohr series of lines of hydrogen spectrum, the third line from the ...
The lines falling in the visible region comprise Balmer series. Hence the third line from red would be n1 =2,
n2 = 5 i.e. 5 → 2.
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In Bohr series of lines of hydrogen spectrum, the third line from the ...
The third line from the red end corresponds to the transition from the fifth energy level to the second energy level. This transition is known as the Balmer series, and it corresponds to the emission of visible light with a wavelength of approximately 656.3 nm.
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In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]Planck’s constant, h = 6.63 × 10–34 Jsa)5 → 2b)4 → 1c)2 → 5d)3 → 2Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]Planck’s constant, h = 6.63 × 10–34 Jsa)5 → 2b)4 → 1c)2 → 5d)3 → 2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]Planck’s constant, h = 6.63 × 10–34 Jsa)5 → 2b)4 → 1c)2 → 5d)3 → 2Correct answer is option 'A'. Can you explain this answer?.
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