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Atoms and Nuclei question

2017 · Q141
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Atoms and Nuclei question

2017 · Q141

NEETPhysicsAtoms and NucleiMCQ+4 / −1
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
  1. A
    1
  2. B
    4
  3. C
    0.5
  4. D
    2
View written solutionFree

Correct answer: B

In case of Balmer series :

n1 = ∞\infty ∞, n2 = 2

∴\therefore∴ 1λB=Rc(122−1∞2){1 \over {{\lambda _B}}} = Rc\left( {{1 \over {{2^2}}} - {1 \over {{\infty ^2}}}} \right)λB​1​=Rc(221​−∞21​) = Rc4{{Rc} \over 4}4Rc​

In case of Lyman series :

n1 = ∞\infty ∞, n2 = 1

∴\therefore∴ 1λL=Rc(112−1∞2){1 \over {{\lambda _L}}} = Rc\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)λL​1​=Rc(121​−∞21​) = Rc

∴\therefore∴ λBλL{{{\lambda _B}} \over {{\lambda _L}}}λL​λB​​ = 41{4 \over 1}14​ = 4

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