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The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is
  • a)
    1215 A°
  • b)
    1640 A°
  • c)
    2430 A°
  • d)
    4687 A°
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The wavelength of the first spectral line in the Balmer series of hydr...
The wave length of first spectral line in the Balmer series of hydrogen atom is 6561Å . Here n2 = 3 and n1 = 2
For the second spectral line in the Balmer series of singly ionised helium ion n2 = 4 and n1 = 2 ; Z = 2
Dividing equation (i) and equation (ii) we get
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The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom isa)1215 A°b)1640 A°c)2430 A°d)4687 A°Correct answer is option 'A'. Can you explain this answer?
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The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom isa)1215 A°b)1640 A°c)2430 A°d)4687 A°Correct 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 The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom isa)1215 A°b)1640 A°c)2430 A°d)4687 A°Correct 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 The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom isa)1215 A°b)1640 A°c)2430 A°d)4687 A°Correct answer is option 'A'. Can you explain this answer?.
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