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Atoms and Nuclei question

2016 · Q137
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Atoms and Nuclei question

2016 · Q137

NEETPhysicsAtoms and NucleiMCQ+4 / −1
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ\lambdaλ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
  1. A
    1625λ{{16} \over {25}}\lambda2516​λ
  2. B
    916λ{9 \over {16}}\lambda169​λ
  3. C
    207λ{{20} \over 7}\lambda720​λ
  4. D
    2013λ{{20} \over {13}}\lambda1320​λ
View written solutionFree

Correct answer: C

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,

1λ=R(1nf2−1ni2){1 \over \lambda } = R\left( {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right)λ1​=R(nf2​1​−ni2​1​)

On jumping from 3rd orbit to 2nd orbit,

1λ=R(122−132){1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)λ1​=R(221​−321​) = 5R36{{5R} \over {36}}365R​

On jumping from 4th orbit to 3rd orbit,

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

∴\therefore∴ λ\lambda λ' = 1447×5λ36{{144} \over 7} \times {{5\lambda } \over {36}}7144​×365λ​ = 20λ7{{20\lambda } \over 7}720λ​

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