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Atoms and Nuclei question

2012 · Q148
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Atoms and Nuclei question

2012 · Q148

NEETPhysicsAtoms and NucleiMCQ+4 / −1
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ\lambdaλ1 : λ\lambdaλ2 emitted in the two cases is
  1. A
    75{7 \over 5}57​
  2. B
    2720{27 \over 20}2027​
  3. C
    275{27 \over 5}527​
  4. D
    207{20 \over 7}720​
View written solutionFree

Correct answer: D

In first case,

n1 = 3 and n2 = 4

1λ1=R(132−142){1 \over {{\lambda _1}}} = R\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right)λ1​1​=R(321​−421​) = 7R144{{7R} \over 144}1447R​

In second case,

n1 = 2 and n2 = 3

1λ2=R(122−132){1 \over {{\lambda _2}}} = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)λ2​1​=R(221​−321​) = 5R36{{5R} \over {36}}365R​

∴\therefore∴ λ1λ2{{{\lambda _1}} \over {{\lambda _2}}}λ2​λ1​​ = 536×1447{5 \over {36}} \times {{144} \over 7}365​×7144​ = 207{20 \over 7}720​

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