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Atoms and Nuclei question

2020 · 9 Jan · Shift 2 · Q52
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Atoms and Nuclei question

2020 · 9 Jan · Shift 2 · Q52

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?
  1. A
    35.8 nm
  2. B
    11.4 nm
  3. C
    8.6 nm
  4. D
    24.2 nm
View written solutionFree

Correct answer: B

  1. Use the ionization energy of a hydrogen-like ion

For a hydrogen-like ion in the ground state, E1=−Z2×1 RydbergE_1 = -Z^2 \times 1\,\text{Rydberg}E1​=−Z2×1Rydberg and the ionization energy from ground state is Z2 Rydbergs.Z^2\,\text{Rydbergs}.Z2Rydbergs.

Given ionization energy = 999 Rydbergs, Z2=9  ⟹  Z=3.Z^2 = 9 \implies Z=3.Z2=9⟹Z=3.

So the ion is a hydrogen-like ion with atomic number 333.

  1. Identify the transition

The second excited state corresponds to n=3n=3n=3 (and ground state is n=1n=1n=1).

So the transition is n=3→n=1.n=3 \to n=1.n=3→n=1.

  1. Find the energy of emitted photon

For a hydrogen-like ion, En=−Z2×13.6n2 eV.E_n = -\frac{Z^2 \times 13.6}{n^2}\,\text{eV}.En​=−n2Z2×13.6​eV.

With Z=3Z=3Z=3: E1=−9×13.6=−122.4 eVE_1 = -9 \times 13.6 = -122.4\,\text{eV}E1​=−9×13.6=−122.4eV E3=−122.49=−13.6 eV.E_3 = -\frac{122.4}{9} = -13.6\,\text{eV}.E3​=−9122.4​=−13.6eV.

Hence photon energy emitted is ΔE=E1−E3\Delta E = E_1 - E_3ΔE=E1​−E3​ in magnitude, ∣ΔE∣=122.4−13.6=108.8 eV.|\Delta E| = 122.4 - 13.6 = 108.8\,\text{eV}.∣ΔE∣=122.4−13.6=108.8eV.

  1. Convert energy to wavelength

Using λ=hcE\lambda = \frac{hc}{E}λ=Ehc​ with hc≈1240 eV⋅nm,hc \approx 1240\,\text{eV·nm},hc≈1240eV⋅nm, we get λ=1240108.8 nm.\lambda = \frac{1240}{108.8}\,\text{nm}.λ=108.81240​nm.

λ≈11.4 nm.\lambda \approx 11.4\,\text{nm}.λ≈11.4nm.

  1. Match with the options

The correct option is B: 11.4 nm.\boxed{\text{B: }11.4\,\text{nm}}.B: 11.4nm​.

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