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Atoms and Nuclei question

2013 · Q138
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Atoms and Nuclei question

2013 · Q138

NEETPhysicsAtoms and NucleiMCQ+4 / −1
Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is
  1. A
    729{7 \over {29}}297​
  2. B
    931{9 \over {31}}319​
  3. C
    527{5 \over {27}}275​
  4. D
    323{3 \over {23}}233​
View written solutionFree

Correct answer: C

We see that wavelength of Lyman series,

nf = 3, ni = 4

1λL=R[132−142]{1 \over {{\lambda _L}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right]λL​1​=R[321​−421​] = 7R144{{7R} \over {144}}1447R​

We see that wavelength of Balmer series :

nf = 2, ni = 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​

Now ratio of longest wavelengths corresponds to Lyman and Balmer series:

λLλB=536×1447{{{\lambda _L}} \over {{\lambda _B}}} = {5 \over {36}} \times {{144} \over 7}λB​λL​​=365​×7144​ = 527{5 \over {27}}275​

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