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Atoms and Nuclei question

2023 · 1 Feb · Shift 2 · Q64
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Atoms and Nuclei question

2023 · 1 Feb · Shift 2 · Q64

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
An electron of a hydrogen like atom, having Z=4Z=4Z=4, jumps from 4th 4^{\text {th }}4th  energy state to 2nd 2^{\text {nd }}2nd  energy state. The energy released in this process, will be : (Given Rch =13.6 eV13.6~\mathrm{eV}13.6 eV) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant
  1. A
    10.5 eV10.5 ~\mathrm{eV}10.5 eV
  2. B
    40.8 eV40.8 ~\mathrm{eV}40.8 eV
  3. C
    13.6 eV13.6 ~\mathrm{eV}13.6 eV
  4. D
    3.4 eV3.4 ~\mathrm{eV}3.4 eV
View written solutionFree

Correct answer: B

  1. Energy levels of a hydrogen-like atom

For a hydrogen-like atom, the energy of the electron in the nthn^{\text{th}}nth orbit is

En=−13.6Z2n2 eVE_n = -\frac{13.6 Z^2}{n^2}\,\text{eV}En​=−n213.6Z2​eV

Given:

  • Z=4Z = 4Z=4
  • Transition: n=4→n=2n=4 \to n=2n=4→n=2

So,

En=−13.6×42n2=−13.6×16n2E_n = -\frac{13.6 \times 4^2}{n^2} = -\frac{13.6 \times 16}{n^2}En​=−n213.6×42​=−n213.6×16​

  1. Find energy of 4th state

E4=−13.6×1642=−13.6×1616=−13.6 eVE_4 = -\frac{13.6 \times 16}{4^2} = -\frac{13.6 \times 16}{16} = -13.6\,\text{eV}E4​=−4213.6×16​=−1613.6×16​=−13.6eV

  1. Find energy of 2nd state

E2=−13.6×1622=−13.6×164=−54.4 eVE_2 = -\frac{13.6 \times 16}{2^2} = -\frac{13.6 \times 16}{4} = -54.4\,\text{eV}E2​=−2213.6×16​=−413.6×16​=−54.4eV

  1. Energy released in the transition

Since the electron goes from a higher level to a lower level, energy released is

ΔE=E4−E2\Delta E = E_4 - E_2ΔE=E4​−E2​

ΔE=(−13.6)−(−54.4)=40.8 eV\Delta E = (-13.6) - (-54.4) = 40.8\,\text{eV}ΔE=(−13.6)−(−54.4)=40.8eV

  1. Check options
  • A: 10.5 eV10.5\,\text{eV}10.5eV ❌
  • B: 40.8 eV40.8\,\text{eV}40.8eV ✅
  • C: 13.6 eV13.6\,\text{eV}13.6eV ❌
  • D: 3.4 eV3.4\,\text{eV}3.4eV ❌

Therefore, the correct answer is Option B.

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