An electron in the hydrogen atom initially in the fourth excited state makes a transition to nth$n^{\text{th}}$ energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________.

Core Logic

Given state:

• Initial state: "fourth excited state"$

  Core Logic

  Given state:

  • Initial state: "fourth excited state" $\Rightarrow n_i = 5
  • Emitted photon energy:$
  • Emitted photon energy: $\Delta E = 2.86\mathrm{~eV}

  Step 1: Calculate energy of initial state ($

Step 1: Calculate energy of initial state ($E_5)

Pattern Recognition

Memorizing the first few energy levels of the hydrogen atom saves calculation time:

• $

  Pattern Recognition

  Memorizing the first few energy levels of the hydrogen atom saves calculation time:

  • $E_1 = -13.6\mathrm{~eV}
  • $
  • $E_2 = -3.4\mathrm{~eV}
  • $
  • $E_3 = -1.51\mathrm{~eV}
  • $
  • $E_4 = -0.85\mathrm{~eV}
  • $
  • $E_5 = -0.54\mathrm{~eV}

  Recognizing that a transition from$

  Recognizing that a transition from $-0.54\mathrm{~eV}by emitting$ by emitting $2.86\mathrm{~eV}lands exactly at$ lands exactly at $-3.4\mathrm{~eV}maps directly to$ maps directly to $n=2$ instantly.

  Chapter Mix

  Class 12 Physics: Atoms