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Atoms and Nuclei question

2015 · Q139
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Atoms and Nuclei question

2015 · Q139

NEETPhysicsAtoms and NucleiMCQ+4 / −1
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
  1. A
    275{{27} \over 5}527​
  2. B
    527{5 \over {27}}275​
  3. C
    49{4 \over 9}94​
  4. D
    94{9 \over 4}49​
View written solutionFree

Correct answer: B

For the longest wavelength in the Lyman series,

n1 = 1 and n2 = 2

1λL=R(112−122){1 \over {{\lambda _L}}} = R\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)λL​1​=R(121​−221​) = 3R4{{3R} \over 4}43R​

For the longest wavelength in the Balmer series,

n1 = 2 and n2 = 3

1λB=R(122−132){1 \over {{\lambda _B}}} = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)λB​1​=R(221​−321​) = 5R36{{5R} \over {36}}365R​

∴\therefore∴ λLλB=43R365R{{{\lambda _L}} \over {{\lambda _B}}} = {{{4 \over {3R}}} \over {{{36} \over {5R}}}}λB​λL​​=5R36​3R4​​ = 527{5 \over {27}}275​

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