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Atoms and Nuclei question

2009 · Q148
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Atoms and Nuclei question

2009 · Q148

NEETPhysicsAtoms and NucleiMCQ+4 / −1
The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths, Maximum wavelength of emitted radiation corresponds to the transition between
  1. A
    n = 3 to n = 1 states
  2. B
    n = 2 to n = 1 states
  3. C
    n = 4 to n = 3 states
  4. D
    n = 3 to n = 2 states
View written solutionFree

Correct answer: C

n(n – 1)/2 = 6

⇒\Rightarrow⇒ n2 – n –12 = 0

⇒\Rightarrow⇒ (n – 4) (n + 3) = 0 or n = 4

So, n = 4 to n = 3 states.

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