Energy Levels — Revision Notes

⚡ 30-Second Revision

Energy of $n$-th orbit (Hydrogen): — $E_n = -\frac{13.6}{n^2} \text{ eV}$
Energy of $n$-th orbit (Hydrogen-like): — $E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$
Radius of $n$-th orbit (Hydrogen-like): — $r_n = \frac{n^2 a_0}{Z}$, where $a_0 = 0.529 \mathring{A}$
Angular Momentum (Bohr's Postulate): — $L = n\frac{h}{2\pi}$
Photon Energy (Transition): — $\Delta E = E_i - E_f = h\nu = \frac{hc}{\lambda}$
Rydberg Formula (Wavelength): — $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$
Ground State: — $n=1$
First Excited State: — $n=2$
Ionization Energy: — Energy to go from $n=1$ to $n=\infty$ (for H, $13.6 \text{ eV}$)
Spectral Series: — Lyman ($n_f=1$, UV), Balmer ($n_f=2$, Visible), Paschen ($n_f=3$, IR)

2-Minute Revision

Energy levels in the Bohr model describe the discrete, quantized energy states an electron can occupy in an atom, particularly hydrogen. These levels are negative, indicating the electron is bound to the nucleus, with $E_n = -13.

6 Z^2/n^2 \text{ eV}$. The principal quantum number$n$dictates the energy and radius ($r_n \propto n^2/Z$). The ground state is$n=1$, and higher$n$values represent excited states. Electrons transition between these levels by absorbing or emitting photons whose energy exactly matches the energy difference ($\Delta E = E_i - E_f$).

This explains atomic line spectra. The Rydberg formula, $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$, relates these transitions to emitted photon wavelengths. Key spectral series include Lyman (to $n=1$, UV), Balmer (to $n=2$, visible), and Paschen (to $n=3$, IR).

Ionization energy is the energy required to remove an electron from the ground state to $n=\infty$, which is $13.6 \text{ eV}$ for hydrogen.

5-Minute Revision

The Bohr model revolutionized our understanding of atomic structure by introducing the concept of quantized energy levels. For a hydrogen atom (or hydrogen-like ions with atomic number $Z$), an electron can only exist in specific, discrete energy states, denoted by the principal quantum number $n=1, 2, 3, \dots$.

The energy of these levels is given by $E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$. The negative sign signifies that the electron is bound to the nucleus, with $0 \text{ eV}$ representing a free electron.

The $n=1$ state is the ground state (most stable), while $n>1$ states are excited states.

Electron transitions between these energy levels are responsible for atomic spectra. When an electron jumps from a higher energy state ($n_i$) to a lower energy state ($n_f$), it emits a photon with energy $\Delta E = E_i - E_f$.

Conversely, absorption of a photon with this exact energy can cause an electron to jump to a higher state. This energy difference is related to the photon's wavelength $\lambda$ by $\Delta E = hc/\lambda$.

The Rydberg formula provides a direct way to calculate the wavelength: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$, where $R$ is the Rydberg constant.

Key Spectral Series for Hydrogen:

Lyman Series: — Transitions to $n_f=1$. These photons are in the ultraviolet (UV) region.
Balmer Series: — Transitions to $n_f=2$. These photons are in the visible region.
Paschen Series: — Transitions to $n_f=3$. These photons are in the infrared (IR) region.

Example: Calculate the energy of the photon emitted when a hydrogen electron drops from $n=3$ to $n=1$. $E_3 = -13.6/3^2 = -1.51 \text{ eV}$ $E_1 = -13.6/1^2 = -13.6 \text{ eV}$ $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV}$. This photon is in the UV region (Lyman series).

Also remember that the radius of the $n$-th orbit is $r_n = \frac{n^2 a_0}{Z}$, where $a_0 = 0.529 \mathring{A}$ is the Bohr radius. The angular momentum is quantized as $L = n\frac{h}{2\pi}$. Ionization energy for hydrogen is $13.6 \text{ eV}$ (from $n=1$ to $n=\infty$).

Prelims Revision Notes

1
Bohr's Postulates:

* Electrons orbit in stable, non-radiating 'stationary states'. * Angular momentum is quantized: $L = mvr = n\frac{h}{2\pi}$. * Energy is emitted/absorbed only during transitions: $h\nu = E_i - E_f$.

1
**Energy Levels ($E_n$):**

* For hydrogen ($Z=1$): $E_n = -\frac{13.6}{n^2} \text{ eV}$. * For hydrogen-like ions (e.g., He$^+$, Li$^{2+}$): $E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$. * Negative sign implies electron is bound. $E=0$ at $n=\infty$ (ionization). * Ground state: $n=1$. First excited state: $n=2$. Second excited state: $n=3$, etc.

1
**Radii of Orbits ($r_n$):**

* For hydrogen-like ions: $r_n = \frac{n^2 a_0}{Z}$, where $a_0 = 0.529 \mathring{A}$ (Bohr radius). * $r_n \propto n^2/Z$.

1
**Velocity of Electron ($v_n$):**

* $v_n \propto Z/n$.

1
Energy Transitions and Photons:

* Emitted photon energy: $\Delta E = E_i - E_f$ (where $E_i > E_f$). * Absorbed photon energy: $\Delta E = E_f - E_i$ (where $E_f > E_i$). * $\Delta E = h\nu = \frac{hc}{\lambda}$.

1
Rydberg Formula for Wavelength:

* $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$, where $R \approx 1.097 \times 10^7 \text{ m}^{-1}$.

1
Spectral Series (Hydrogen):

* Lyman Series: $n_f=1$, $n_i=2,3,4,\dots$. UV region. Max energy transition is $n=\infty \to n=1$ (ionization energy). * Balmer Series: $n_f=2$, $n_i=3,4,5,\dots$. Visible region. First line $H_\alpha$ is $n=3 \to n=2$. * Paschen Series: $n_f=3$, $n_i=4,5,6,\dots$. Infrared (IR) region. * Brackett ($n_f=4$) and Pfund ($n_f=5$) are also IR.

1
Ionization Energy: — Energy required to remove an electron from its ground state ($n=1$) to $n=\infty$. For hydrogen, $13.6 \text{ eV}$.
2
Relationship between KE, PE, and E: — For a Bohr orbit, $KE = -E$ and $PE = 2E$. Also, $PE = -2KE$.

Vyyuha Quick Recall

To remember the order of spectral series and their regions: Lazy Boys Play Baseball Professionally Lyman (n=1) - UltraViolet Balmer (n=2) - Visible Paschen (n=3) - InfraRed Brackett (n=4) - InfraRed Pfund (n=5) - InfraRed (Remember UV, Visible, IR for the first three, then all others are IR.)