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JEE Mains · Physics · STD 12 - 12. atoms

In \(Li^{++},\) electron in first Bohr orbit is excited to a level by a radiation of wavelength \(\lambda .\) When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of \(\lambda \)?.....\(nm\) (Given: \(h = 6.63\times 10^{34}\,js; e = 3 \times 10^8\,ms^{-1}\) )

  1. A \(10.8\)
  2. B \(11.4\)
  3. C \(9.4\)
  4. D \(12.3\)
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Answer & Solution

Correct Answer

(A) \(10.8\)

Step-by-step Solution

Detailed explanation

\(\frac{h c}{\lambda}=13.6\, \mathrm{eV}(\mathrm{g})\left\{1-\frac{1}{16}\right\}\) \(\frac{1240\, \mathrm{eV}}{\lambda}=\frac{15}{16} \times 9 \times 13.6\, \mathrm{eV}\) \(\lambda=\frac{1240 \times 16}{15 \times 9 \times 13.6}=10.8\, \mathrm{nm}\)
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