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Atoms and Nuclei question

2006 · Q148
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Atoms and Nuclei question

2006 · Q148

NEETPhysicsAtoms and NucleiMCQ+4 / −1
Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emited by hydrogen will be
  1. A
    one
  2. B
    two
  3. C
    three
  4. D
    four
View written solutionFree

Correct answer: C

Energy required for exciting the hydrogen atom in the ground state to orbit n is given by

E = En - E1

⇒\Rightarrow⇒ 12.1 = −13.6n2 - {{13.6} \over {{n^2}}}−n213.6​ - (−13.612)\left( { - {{13.6} \over {{1^2}}}} \right)(−1213.6​) = 3

⇒\Rightarrow⇒ n2 = 13.61.5{{{13.6} \over {1.5}}}1.513.6​ = 9

⇒\Rightarrow⇒ n = 3

Number of spectral lines emitted

= n(n−1)2{{n\left( {n - 1} \right)} \over 2}2n(n−1)​ = 3×22{{3 \times 2} \over 2}23×2​

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