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For the Balmer series in the spectrum of H atom,  the correct statements among (l) to (IV) are:
(I) As wavelength decreases, the lines in the series converge.
(II) The integer n1 is equal to 2.
(III) The lines of longest wavelength correspond to n2 = 3.
(IV) The ionisation energy of hydrogen can be calculated from wave number of these lines.
  • a)
    (I), (II), (III) 
  • b)
    (II), (III), (IV)
  • c)
    (I), (III), (IV)
  • d)
    (I), (II), (IV)
Correct answer is option 'A'. Can you explain this answer?
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For the Balmer series in the spectrum of H atom,the correct statements...
In the Balmer series of H-atom, electronic transitions take place from higher orbits to 2nd orbit, and the longest wavelength will correspond to the transition from 3rd orbit to 2nd orbit.
∴ n1 = 2 and n2 = 3 for longest wavelength.
As wavelength decreases, the lines in the Balmer series converge. The correct statements are (I), (II) and (III).
The ionisation energy of hydrogen can be calculated from wave number of the last line of the Lyman series:
n1 = 1 and n2 = ∞
Hence, statement IV is incorrect.
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For the Balmer series in the spectrum of H atom,the correct statements among (l) to (IV) are:(I) As wavelength decreases, the lines in the series converge.(II) The integer n1 is equal to 2.(III) The lines of longest wavelength correspond to n2 = 3.(IV) The ionisation energy of hydrogen can be calculated from wave number of these lines.a)(I), (II), (III)b)(II), (III), (IV)c)(I), (III), (IV)d)(I), (II), (IV)Correct answer is option 'A'. Can you explain this answer?
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