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Atoms and Nuclei question

2016 · Q136
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Atoms and Nuclei question

2016 · Q136

NEETPhysicsAtoms and NucleiMCQ+4 / −1
Given the value of Rydberg constant is 107 m−-−1, the wave number of the last line of the Balmer series in hydrogen spectrum will be
  1. A
    0.25 ×\times× 107 m−-−1
  2. B
    2.5 ×\times× 107 m−-−1
  3. C
    0.025 ×\times× 104 m−-−1
  4. D
    0.5 ×\times× 107 m−-−1
View written solutionFree

Correct answer: A

The wave number of the last line of the Balmer series in hydrogen spectrum is given by

1λ=R(122−1∞2){1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{\infty ^2}}}} \right)λ1​=R(221​−∞21​)

= R4{R \over 4}4R​ = 1074{{{{10}^7}} \over 4}4107​ = 0.25 ×\times× 107 m−-−1

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