Related Formula
The energy of an electron in the n$n$-th shell of a hydrogen atom is:
Eₙ = -(13.6)/(n²)~eV$$E_n = -\frac{13.6}{n^2}\mathrm{~eV}$$
The emitted photon energy during transition nᵢ arrow nf$n_i \rightarrow n_f$ is:
Δ E = Enᵢ - Enf$$\Delta E = E_{n_i} - E_{n_f}$$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
)$)
$E₅ = -(13.6)/(5²) = -(13.6)/(25) = -0.544~eV$E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} = -0.544\mathrm{~eV}$Step 2: Calculate final state energy ($Eₙ$)
$$
Step 2: Calculate final state energy ($E_n$)
$$2.86 = E_5 - E_n = -0.544 - E_n
$$
$$E_n = -0.544 - 2.86 = -3.404\mathrm{~eV}
$ Step 3: Determine the integer shell index ($n$)
$$$
Step 3: Determine the integer shell index ($n$)
$$-3.4 = -\frac{13.6}{n^2}
$$
$$n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2$
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