Atoms — Part 2: Energy levels and the hydrogen spectrum: visual lesson, ISC Class 12 Physics

Lesson

Energy Levels and the Hydrogen Spectrum

Part 2 — why the electron’s energy is negative, and how jumps between levels write a barcode of lines.

In this lesson you will

Derive the kinetic, potential and total energy of the electron in the nth orbit and explain why the total energy is negative
Draw the energy-level diagram of hydrogen and identify the Lyman, Balmer, Paschen, Brackett and Pfund series
Calculate the wavelength of a line from a transition using the Rydberg formula, and identify its series and region

Concept

Kinetic, potential and total energy

From $\dfrac{mv^2}{r} = \dfrac{e^2}{4\pi\varepsilon_0 r^2}$ : $K = \dfrac{1}{2}mv^2 = \dfrac{e^2}{8\pi\varepsilon_0 r}$ . The potential energy of the electron (−e) at distance r from the proton (+e) is $U = -\dfrac{e^2}{4\pi\varepsilon_0 r}$ . So $E = K + U = -\dfrac{e^2}{8\pi\varepsilon_0 r},\qquad K = -E,\quad U = 2E$

Table of the kinetic, potential and total energy of the electron in a Bohr orbit, in terms of the radius and of n
Energy	In terms of r	In terms of n
Kinetic K	$+\dfrac{e^2}{8\pi\varepsilon_0 r}$	$+\dfrac{13.6}{n^2}$ eV
Potential U	$-\dfrac{e^2}{4\pi\varepsilon_0 r}$	$-\dfrac{27.2}{n^2}$ eV
Total E = K + U	$-\dfrac{e^2}{8\pi\varepsilon_0 r}$	$-\dfrac{13.6}{n^2}$ eV

Concept

Energy of the nth orbit

Put $r_n = \dfrac{\varepsilon_0 n^2 h^2}{\pi m e^2}$ into $E = -\dfrac{e^2}{8\pi\varepsilon_0 r}$ : $E_n = -\dfrac{me^4}{8\varepsilon_0^2 n^2 h^2} = -\dfrac{13.6}{n^2}\ \text{eV}$

E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.85 eV, …, E∞ = 0.
As n increases, E increases (becomes less negative) while K decreases.

kinetic energy 13.6/n²
potential energy −27.2/n²
total energy −13.6/n²

K = −E and U = 2E in every orbit; all three tend to zero as n → ∞.

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