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Structure of Atom question

2017 · 9 Apr · Shift 1 · Q21
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Structure of Atom question

2017 · 9 Apr · Shift 1 · Q21

JEE MainChemistryStructure of AtomMCQ+4 / −1
The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with :
  1. A
    Lyman series
  2. B
    Balmer series
  3. C
    Paschen series
  4. D
    Brackett series
View written solutionFree

Correct answer: B

  1. Use Bohr radius relation for hydrogen

For hydrogen atom, rn=n2a0r_n = n^2 a_0rn​=n2a0​ where a0=52.9 pma_0 = 52.9\,\text{pm}a0​=52.9pm

  1. Given radius

The electron falls to an orbit of radius rn=211.6 pmr_n = 211.6\,\text{pm}rn​=211.6pm

So, n2a0=211.6n^2 a_0 = 211.6n2a0​=211.6 n2=211.652.9=4n^2 = \frac{211.6}{52.9} = 4n2=52.9211.6​=4 n=2n=2n=2

Thus, the transition ends at the orbit with principal quantum number nf=2n_f = 2nf​=2

  1. Identify the spectral series

In hydrogen spectrum:

  • Lyman series: transitions ending at n=1n=1n=1
  • Balmer series: transitions ending at n=2n=2n=2
  • Paschen series: transitions ending at n=3n=3n=3
  • Brackett series: transitions ending at n=4n=4n=4

Since the transition ends at nf=2n_f=2nf​=2 it belongs to the Balmer series.

  1. Check options
  • A: Lyman series →\to→ ends at n=1n=1n=1, incorrect
  • B: Balmer series →\to→ ends at n=2n=2n=2, correct
  • C: Paschen series →\to→ ends at n=3n=3n=3, incorrect
  • D: Brackett series →\to→ ends at n=4n=4n=4, incorrect

Therefore, the correct answer is B.

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