Hydrogen Spectrum and Electronic Configuration
Bohr's Model of the Hydrogen Atom
Building on Planck quantum theory, Bohr proposed that the electron in a hydrogen atom can only exist in certain permitted quantized energy levels. The Bohr model rests on four postulates: the electron revolves in fixed circular orbits each assigned a quantum number; the electron neither emits nor absorbs energy while in the same orbit; energy is emitted or absorbed only when the electron jumps between orbits, with the energy change governed by Planck equation; and the angular momentum of the electron is quantized in integer multiples of $h/2\pi$.
$$\Delta E = E_2 - E_1 = h\nu$$
Energy difference between two orbits equals the energy of the photon absorbed or emitted during the transition
$\Delta E$=energy difference between two orbits(J)
$E_2$=energy of the higher orbit(J)
$E_1$=energy of the lower orbit(J)
$h$=Planck constant, $6.626 \times 10^{-34}$ J·s(J·s)
$\nu$=frequency of the photon(Hz (s$^{-1}$))
$E_2 > E_1$
→electron absorbs energy (jumps to higher orbit)
$E_2 < E_1$
→electron emits energy (falls to lower orbit)
Quantized Orbits: Each orbit has a fixed energy and is assigned a principal quantum number $n = 1, 2, 3, \ldots$. The electron is stable only in these orbits, never in between.
Stationary State: While revolving in a fixed orbit, the electron does not radiate energy — this overcomes the catastrophic prediction of Rutherford's model where orbiting electrons should continuously lose energy.
Angular Momentum Quantization: The angular momentum $mvr$ of the electron equals $nh/2\pi$, meaning it can only take values $h/2\pi, 2h/2\pi, 3h/2\pi, \ldots$ but nothing in between.
Bohr derived the Bohr radius of the $n$th orbit by balancing the electrostatic force of attraction between the nucleus and electron against the centrifugal force of the revolving electron, then substituting the quantized angular momentum condition to eliminate velocity.
$$r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}$$
Radius of the nth orbit of a hydrogen-like atom, showing direct proportionality to $n^2$ and inverse proportionality to $Z$
$r_n$=radius of the nth orbit(m)
$\varepsilon_0$=vacuum permittivity, $8.85 \times 10^{-12}$ C$^2$J$^{-1}$m$^{-1}$(C$^2$J$^{-1}$m$^{-1}$)
$n$=principal quantum number(dimensionless)
$h$=Planck constant, $6.626 \times 10^{-34}$ J·s(J·s)
$m$=mass of electron, $9.109 \times 10^{-31}$ kg(kg)
$Z$=proton number (1 for hydrogen)(dimensionless)
$e$=charge on electron, $1.602 \times 10^{-19}$ C(C)
$Z = 1$ (hydrogen)
→$r_n = 0.529 \times 10^{-10} \, n^2$ m $= 0.529 \, n^2$ Å
$n = 1$
→$r_1 = 0.529$ Å, called the Bohr radius
Force Balance: The electrostatic force $Ze^2/(4\pi\varepsilon_0 r^2)$ equals the centrifugal force $mv^2/r$, giving $mv^2 = Ze^2/(4\pi\varepsilon_0 r)$.
Eliminating Velocity: Substituting $v = nh/(2\pi mr)$ from the angular momentum condition and solving for $r$ yields the Bohr radius formula.
Radius Scaling: For hydrogen ($Z=1$), $r_n = 0.529 \, n^2$ Å. The 2nd orbit is four times farther, the 3rd is nine times farther, and the 4th is sixteen times farther than the 1st orbit.
Orbital Spacing: The gaps between orbits increase with $n$: $r_2 - r_1 < r_3 - r_2 < r_4 - r_3$, so orbits are not equally spaced.
Radii of Hydrogen Atom Orbits
1
$n=1$: $r_1 = 0.529$ Å
2
$n=2$: $r_2 = 2.11$ Å
3
$n=3$: $r_3 = 4.75$ Å
4
$n=4$: $r_4 = 8.40$ Å
5
$n=5$: $r_5 = 13.22$ Å
The energy level of the electron in the $n$th orbit is derived from the sum of kinetic and potential energy. The total energy comes out negative, reflecting that the electron is bound to the nucleus. As $n$ increases, energy becomes less negative (closer to zero), meaning the electron is less tightly held.
$$E_n = -2.178 \times 10^{-18} \left[ \frac{1}{n^2} \right] \text{ J}$$
Energy of the electron in the nth orbit of hydrogen, inversely proportional to $n^2$
$E_n$=energy of electron in nth orbit(J per atom)
$2.178 \times 10^{-18}$=combined constant $me^4/(8\varepsilon_0^2 h^2)$(J)
$n$=principal quantum number(dimensionless)
$n = \infty$
→$E_\infty = 0$, the electron is free from the nucleus
$n = 1$
→$E_1 = -2.178 \times 10^{-18}$ J, the most tightly bound state (ground state)
Derivation: Starting from $E = \frac{1}{2}mv^2 - \frac{Ze^2}{4\pi\varepsilon_0 r}$, substituting $mv^2 = Ze^2/(4\pi\varepsilon_0 r)$ from the force balance gives $E = -\frac{Ze^2}{8\pi\varepsilon_0 r}$, then replacing $r$ with the Bohr radius expression.
Negative Sign: The negative value means work must be done to remove the electron from the atom. At infinity, $E = 0$ and the electron is free.
Energy in kJ/mol: Multiplying by Avogadro's number and dividing by 1000 gives $E_n = -1313.31/n^2$ kJ·mol$^{-1}$.
Ionization Energy: The energy difference $E_\infty - E_1 = 0 - (-1313.31) = 1313.31$ kJ·mol$^{-1}$ equals the ionization energy of hydrogen.
Energy of Electron in Hydrogen Atom Orbits
1
$n=1$: $E_1 = -1313.31$ kJ·mol$^{-1}$
2
$n=2$: $E_2 = -328.32$ kJ·mol$^{-1}$
3
$n=3$: $E_3 = -145.92$ kJ·mol$^{-1}$
4
$n=4$: $E_4 = -82.08$ kJ·mol$^{-1}$
5
$n=5$: $E_5 = -52.53$ kJ·mol$^{-1}$
6
$n=\infty$: $E_\infty = 0$ kJ·mol$^{-1}$ (free electron)
Hydrogen Spectrum
When radiation of light passes through a prism, different wavelengths bend by different amounts. White light splits into a continuous band of colours from violet (400 nm) to red (750 nm). A continuous spectrum has no sharp boundary between colours, like a rainbow from sunlight. In contrast, a line spectrum consists of distinct bright lines separated by dark spaces, characteristic of individual elements. The hydrogen line spectrum shows several isolated sharp lines in the visible, ultraviolet, and infrared regions.
Continuous Spectrum: Produced by incandescent solids or the sun. Colours merge smoothly into one another with no dark gaps. This is a property of matter in bulk.
Line Spectrum: Produced when an element or its compound is volatilized in a flame or when gaseous elements are heated or subjected to electric discharge. Each element has a unique set of lines.
Emission vs Absorption: In emission spectrum, bright lines appear against a dark background (light emitted by excited atoms). In absorption spectrum, dark lines appear against a bright background (specific wavelengths absorbed from white light). The wavelengths of corresponding lines are identical in both.
The spectral lines of hydrogen are classified into five spectral series named after their discoverers. Each series arises when the electron falls to a specific lower energy level. The Lyman series (ultraviolet) results from transitions to $n=1$, the Balmer series (visible) from transitions to $n=2$, and the Paschen, Brackett, and Pfund series (all infrared) from transitions to $n=3$, $n=4$, and $n=5$ respectively.
Spectral Series of Hydrogen
1
Lyman series: transitions to $n_1 = 1$ (UV region)
2
Balmer series: transitions to $n_1 = 2$ (visible region); lines named $H_\alpha, H_\beta, H_\gamma, H_\delta, \ldots$
3
Paschen series: transitions to $n_1 = 3$ (IR region)
4
Brackett series: transitions to $n_1 = 4$ (IR region)
5
Pfund series: transitions to $n_1 = 5$ (IR region)
Wave Numbers of Balmer Series Lines
1
$H_\alpha$: $15.21 \times 10^5$ m$^{-1}$ ($n_2 = 3 \to n_1 = 2$)
2
$H_\beta$: $20.60 \times 10^5$ m$^{-1}$ ($n_2 = 4 \to n_1 = 2$)
3
$H_\gamma$: $23.5 \times 10^5$ m$^{-1}$ ($n_2 = 5 \to n_1 = 2$)
4
$H_\delta$: $24.35 \times 10^5$ m$^{-1}$ ($n_2 = 6 \to n_1 = 2$)
The wave number (reciprocal of wavelength) of any spectral line in the hydrogen spectrum is given by the Rydberg formula, derived from Bohr's energy equation combined with $E = h\nu$ and $\nu = c\bar{\nu}$. The Rydberg constant $R_H = 1.09678 \times 10^7$ m$^{-1}$ is the combined value of all physical constants in the equation.
$$\bar{\nu} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] m^{-1}$$
Wave number of a photon emitted when an electron transitions between two energy levels in hydrogen
$\bar{\nu}$=wave number of the spectral line(m$^{-1}$)
$R_H$=Rydberg constant for hydrogen, $1.09678 \times 10^7$(m$^{-1}$)
$n_1$=lower energy level (final orbit)(dimensionless)
$n_2$=higher energy level (initial orbit)(dimensionless)
$n_1 = 1$, $n_2 = 2$
→first line of Lyman series, $\bar{\nu} = 82.26 \times 10^5$ m$^{-1}$
$n_1 = 2$, $n_2 = \infty$
→limiting line of Balmer series, $\bar{\nu} = 27.42 \times 10^5$ m$^{-1}$
Energy-Wavelength Link: The energy difference $\Delta E = E_2 - E_1$ determines both the frequency $\nu = \Delta E / h$ and wave number $\bar{\nu} = \Delta E / (hc)$ of the emitted or absorbed photon.
Limiting Line: When $n_2 \to \infty$, the term $1/n_2^2 \to 0$ and $\bar{\nu}_{\text{limit}} = R_H / n_1^2$. This is the maximum wave number (shortest wavelength) of the series.
Decreasing Wave Numbers: As we go from Lyman to Pfund series, the wave numbers of corresponding lines decrease because the energy gaps between levels shrink with increasing $n$.
Despite its success with hydrogen, the Bohr model has several important limitations. It explains spectra only for one-electron systems like H, $\text{He}^+$, $\text{Li}^{2+}$, and $\text{Be}^{3+}$, but fails for multi-electron atoms. High-resolution spectroscopy reveals fine structure — each spectral line is actually a cluster of closely spaced lines, suggesting that one quantum number alone cannot fully describe the electron. Bohr's model also cannot explain the Zeeman effect (splitting of lines in a magnetic field) or the Stark effect (splitting in an electric field), and the assumption of circular orbits in a single plane is incorrect — electrons move in three-dimensional space.
One-Electron Limitation: The model works only for hydrogen and hydrogen-like ions. It cannot predict spectra of helium, lithium, beryllium, or any multi-electron atom.
Fine Structure: The $H_\alpha$ line in the Balmer series actually consists of five closely spaced component lines, not one. This indicates additional quantum numbers are needed.
Three-Dimensional Motion: Sommerfeld (1915) extended Bohr's model by proposing elliptical orbits with the nucleus at one focus, partially addressing this limitation.
Zeeman and Stark Effects: Neither the splitting of spectral lines in magnetic fields (Zeeman) nor in electric fields (Stark) can be explained by Bohr's theory alone.
Electronic Configuration Rules
Before applying the filling rules, subshells must be arranged in order of increasing energy using the n+l rule. Subshells with smaller $n+l$ values are lower in energy. When two subshells have the same $n+l$ value, the one with the smaller $n$ is filled first. This ordering explains why 4s fills before 3d, and why the energy sequence is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s.
Rule Statement: Arrange subshells in increasing order of $(n+l)$; if $(n+l)$ is the same, the subshell with lower $n$ comes first.
Key Inversions: $4s$ ($n+l=4$) fills before $3d$ ($n+l=5$); $5s$ ($n+l=5$) fills before $4d$ ($n+l=6$); $6s$ ($n+l=6$) fills before $4f$ ($n+l=7$).
Degenerate Orbitals: Orbitals in the same subshell (e.g., $p_x, p_y, p_z$) have the same energy — they are called degenerate orbitals. The $(n+l)$ rule does not apply within a subshell.
Subshell Ordering by (n+l) Rule
1
1s ($1+0=1$), 2s ($2+0=2$), 2p ($2+1=3$), 3s ($3+0=3$)
2
3p ($3+1=4$), 4s ($4+0=4$), 3d ($3+2=5$), 4p ($4+1=5$)
3
5s ($5+0=5$), 4d ($4+2=6$), 5p ($5+1=6$), 6s ($6+0=6$)
4
4f ($4+3=7$), 5d ($5+2=7$), 6p ($6+1=7$), 7s ($7+0=7$)
The Aufbau principle (German for 'building up') states that electrons fill subshells in order of increasing energy. Starting from the lowest energy subshell (1s), each subshell must be completely filled before electrons enter the next higher energy subshell. This principle, combined with the n+l rule ordering, determines the ground-state electronic configuration of any atom.
Building Order: Electrons occupy 1s first, then 2s, then 2p, then 3s, then 3p, then 4s, then 3d, and so on following the energy sequence.
Maximum Electrons per Subshell: Each subshell holds $2(2\ell+1)$ electrons: s holds 2, p holds 6, d holds 10, f holds 14.
Maximum per Shell: The maximum number of electrons in shell $n$ is $2n^2$. For $n=1$: 2 electrons; $n=2$: 8; $n=3$: 18; $n=4$: 32.
Notation: Electron configurations are written using noble gas cores in brackets for brevity. For example, sodium ($Z=11$) is $[\text{Ne}]\,3s^1$ instead of $1s^2 2s^2 2p^6 3s^1$.
The Pauli exclusion principle states that no two electrons in the same atom can have identical values for all four quantum numbers. In practical terms, this means an orbital can hold at most two electrons, and those two must have opposite spins. If one electron has spin $+\frac{1}{2}$ (↑), the other must have spin $-\frac{1}{2}$ (↓).
Statement: Two electrons in the same orbital must differ in at least one quantum number. Since $n$, $\ell$, and $m_\ell$ are identical for electrons in the same orbital, the spin quantum number $m_s$ must be different.
Spin Pairing: The two electrons in an orbital are said to be paired (↑↓). A single electron in an orbital is unpaired (↑).
Consequence for Maximum Occupancy: An orbital holds at most 2 electrons, a subshell holds at most $2(2\ell+1)$ electrons, and a shell holds at most $2n^2$ electrons — all direct consequences of Pauli's principle.
Helium Example: $1s^2$ has two electrons with spins ↑↓ — both share $n=1, \ell=0, m_\ell=0$ but differ in $m_s$.
Hund rule states that when electrons are placed in degenerate orbitals (orbitals of equal energy within the same subshell), they occupy separate orbitals with parallel (same-direction) spins before any pairing occurs. This minimizes electron-electron repulsion and produces the most stable arrangement.
Rule: In a subshell with multiple degenerate orbitals (e.g., three 2p orbitals), each orbital gets one electron with the same spin before any orbital receives a second electron.
Carbon Example: $_6\text{C} = 1s^2 \, 2s^2 \, 2p_x^1 \, 2p_y^1 \, 2p_z^0$ — the two 2p electrons occupy separate orbitals with parallel spins rather than pairing in one orbital.
Nitrogen Example: $_7\text{N} = 1s^2 \, 2s^2 \, 2p_x^1 \, 2p_y^1 \, 2p_z^1$ — three unpaired electrons, all with parallel spins. Nitrogen has maximum unpaired electrons in its valence shell.
Oxygen Example: $_8\text{O} = 1s^2 \, 2s^2 \, 2p_x^2 \, 2p_y^1 \, 2p_z^1$ — the fourth electron must pair in $2p_x$ since all three orbitals have one electron each.
Stability Reason: Parallel spins keep electrons farther apart on average, reducing electrostatic repulsion.
Certain elements deviate from the expected Aufbau filling because a half-filled or fully filled d-subshell provides extra stability. The most notable exceptions are chromium ($Z=24$) and copper ($Z=29$), where one electron from the 4s subshell moves to the 3d subshell to achieve a more stable configuration.
Chromium ($Z=24$): Expected $[\text{Ar}]\,4s^2\,3d^4$, actual $[\text{Ar}]\,4s^1\,3d^5$. The half-filled $3d^5$ subshell is more stable than $3d^4$.
Copper ($Z=29$): Expected $[\text{Ar}]\,4s^2\,3d^9$, actual $[\text{Ar}]\,4s^1\,3d^{10}$. The fully filled $3d^{10}$ subshell is more stable than $3d^9$.
Stability Pattern: Half-filled ($d^5, f^7$) and fully filled ($d^{10}, f^{14}$) subshells have lower energy due to symmetrical electron distribution and exchange energy.
Applying the Aufbau principle, Pauli exclusion principle, and Hund rule together yields the ground-state electronic configuration for every element. The table below shows configurations for the first 30 elements, illustrating how the three rules work in practice.
Electronic Configurations of Elements (Z = 1–20)
1
$_1\text{H}$: $1s^1$
2
$_2\text{He}$: $1s^2$
3
$_3\text{Li}$: $[\text{He}]\,2s^1$
4
$_4\text{Be}$: $[\text{He}]\,2s^2$
5
$_5\text{B}$: $[\text{He}]\,2s^2\,2p^1$
6
$_6\text{C}$: $[\text{He}]\,2s^2\,2p^2$ ($2p_x^1\,2p_y^1$ by Hund's rule)
7
$_7\text{N}$: $[\text{He}]\,2s^2\,2p^3$ ($2p_x^1\,2p_y^1\,2p_z^1$, max unpaired)
8
$_8\text{O}$: $[\text{He}]\,2s^2\,2p^4$ (pairing begins in $2p$)
9
$_9\text{F}$: $[\text{He}]\,2s^2\,2p^5$
10
$_{10}\text{Ne}$: $[\text{He}]\,2s^2\,2p^6$ (noble gas core)
11
$_{11}\text{Na}$: $[\text{Ne}]\,3s^1$
12
$_{12}\text{Mg}$: $[\text{Ne}]\,3s^2$
13
$_{13}\text{Al}$: $[\text{Ne}]\,3s^2\,3p^1$
14
$_{14}\text{Si}$: $[\text{Ne}]\,3s^2\,3p^2$
15
$_{15}\text{P}$: $[\text{Ne}]\,3s^2\,3p^3$
16
$_{16}\text{S}$: $[\text{Ne}]\,3s^2\,3p^4$
17
$_{17}\text{Cl}$: $[\text{Ne}]\,3s^2\,3p^5$
18
$_{18}\text{Ar}$: $[\text{Ne}]\,3s^2\,3p^6$ (noble gas core)
19
$_{19}\text{K}$: $[\text{Ar}]\,4s^1$
20
$_{20}\text{Ca}$: $[\text{Ar}]\,4s^2$
Electronic Configurations of Elements (Z = 21–30, Transition Series)
1
$_{21}\text{Sc}$: $[\text{Ar}]\,4s^2\,3d^1$
2
$_{22}\text{Ti}$: $[\text{Ar}]\,4s^2\,3d^2$
3
$_{23}\text{V}$: $[\text{Ar}]\,4s^2\,3d^3$
4
$_{24}\text{Cr}$: $[\text{Ar}]\,4s^1\,3d^5$ (exception: half-filled $d$)
5
$_{25}\text{Mn}$: $[\text{Ar}]\,4s^2\,3d^5$
6
$_{26}\text{Fe}$: $[\text{Ar}]\,4s^2\,3d^6$
7
$_{27}\text{Co}$: $[\text{Ar}]\,4s^2\,3d^7$
8
$_{28}\text{Ni}$: $[\text{Ar}]\,4s^2\,3d^8$
9
$_{29}\text{Cu}$: $[\text{Ar}]\,4s^1\,3d^{10}$ (exception: full $d$)
10
$_{30}\text{Zn}$: $[\text{Ar}]\,4s^2\,3d^{10}$
Application Strategy
Worked Examples
Problem Statement
Calculate the radius of the third orbit of the hydrogen atom.
Bohr radius
Strategic Toolkit
$$r_n = 0.529 \, n^2 \text{ Å}$$
Step 1
Identify n and substitute
For the third orbit, n = 3. Substitute into the hydrogen-specific Bohr radius formula.
$$r_3 = 0.529 \times 3^2 \text{ Å}$$
Step 2
Calculate
Compute the square and multiply by the Bohr radius constant.
$$r_3 = 0.529 \times 9 = 4.76 \text{ Å}$$
Problem Statement
Calculate the energy of an electron in the 2nd orbit of the hydrogen atom in kJ·mol$^{-1}$.
energy level
Strategic Toolkit
$$E_n = -\frac{1313.31}{n^2} \text{ kJmol}^{-1}$$
Step 1
Substitute n = 2
Use the kJ·mol$^{-1}$ form of the Bohr energy equation for the 2nd orbit.
$$E_2 = -\frac{1313.31}{2^2} \text{ kJmol}^{-1}$$
Step 2
Calculate
Divide by 4 to get the energy value.
$$E_2 = -\frac{1313.31}{4} = -328.33 \text{ kJmol}^{-1}$$
Problem Statement
Calculate the wave number of the first line ($H_\alpha$) of the Balmer series of hydrogen.
Rydberg formulaBalmer series
Strategic Toolkit
$$\bar{\nu} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$$
Step 1
Identify transitions
The first Balmer line corresponds to the transition $n_2 = 3$ to $n_1 = 2$.
$$\bar{\nu} = 1.09678 \times 10^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \text{ m}^{-1}$$
Step 2
Evaluate the bracket
Calculate the difference of the reciprocal squares.
$$\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} = 0.1389$$
Step 3
Multiply by Rydberg constant
Multiply the Rydberg constant by the bracket value to get the wave number.
$$\bar{\nu} = 1.09678 \times 10^7 \times 0.1389 = 15.23 \times 10^5 \text{ m}^{-1}$$
Problem Statement
Calculate the energy difference when an electron jumps from $n=3$ to $n=1$ in a hydrogen atom. Express the answer in joules per atom.
quantized energy level
Strategic Toolkit
$$\Delta E = E_3 - E_1 = 2.178 \times 10^{-18} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \text{ J}$$
Step 1
Set up the energy difference equation
The electron falls from n=3 (higher) to n=1 (lower), so it emits energy. The magnitude of the energy difference is calculated using the Bohr equation with $n_1 = 1$ and $n_2 = 3$.
$$\Delta E = 2.178 \times 10^{-18} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \text{ J}$$
Step 2
Evaluate the bracket
Compute the difference of reciprocal squares.
$$1 - \frac{1}{9} = \frac{8}{9}$$
Step 3
Calculate the energy
Multiply the constant by the bracket value.
$$\Delta E = 2.178 \times 10^{-18} \times \frac{8}{9} = 1.936 \times 10^{-18} \text{ J}$$
Problem Statement
Calculate the wave number of the limiting line of the Lyman series of hydrogen.
Rydberg formulaLyman series
Strategic Toolkit
$$\bar{\nu}_{\text{limit}} = \frac{R_H}{n_1^2}$$
Step 1
Set up for limiting line
The limiting line occurs when the electron falls from $n_2 = \infty$ to $n_1 = 1$. Since $1/\infty^2 = 0$, the wave number simplifies.
$$\bar{\nu} = 1.09678 \times 10^7 \left[ \frac{1}{1^2} - 0 \right] \text{ m}^{-1}$$
Step 2
Calculate
The wave number equals the Rydberg constant.
$$\bar{\nu} = 1.09678 \times 10^7 = 109.678 \times 10^5 \text{ m}^{-1}$$
Problem Statement
An electron in a hydrogen atom has energy $-0.242 \times 10^{-18}$ J. Calculate the principal quantum number $n$ of this orbit.
energy level
Strategic Toolkit
$$E_n = -\frac{2.178 \times 10^{-18}}{n^2} \text{ J}$$
Step 1
Set up the equation
Equate the given energy to the Bohr energy formula and solve for $n$.
$$-0.242 \times 10^{-18} = -\frac{2.178 \times 10^{-18}}{n^2}$$
Step 2
Rearrange for $n^2$
Cancel negative signs and the $10^{-18}$ factor, then solve for $n^2$.
$$n^2 = \frac{2.178}{0.242} = 9.0$$
Step 3
Solve for n
Take the square root to find the principal quantum number.
$$n = \sqrt{9.0} = 3$$
Problem Statement
An electron in hydrogen jumps from $n=5$ to $n=2$. Identify the spectral series and the spectral region of the emitted photon.
spectral seriesBalmer series
Strategic Toolkit
$$n_1 = 2 \Rightarrow \text{Balmer series}$$
Step 1
Identify the lower level
The electron falls to $n_1 = 2$, which identifies the Balmer series.
$$n_1 = 2 \Rightarrow \text{Balmer series (visible region)}$$
Step 2
Verify with wave number
Calculate the wave number and confirm it falls in the Balmer range.
$$\bar{\nu} = 1.09678 \times 10^7 \left[ \frac{1}{4} - \frac{1}{25} \right] = 23.04 \times 10^5 \text{ m}^{-1}$$
Problem Statement
Write the ground-state electronic configuration of chromium ($Z=24$), explaining any deviation from the expected Aufbau order.
Aufbau principlePauli exclusion principleHund rule
Strategic Toolkit
$$_{24}\text{Cr} = [\text{Ar}]\,4s^1\,3d^5$$
Step 1
Expected configuration by Aufbau
Following strict Aufbau order, 4s fills before 3d, giving 4s$^2$3d$^4$.
$$\text{Expected: } [\text{Ar}]\,4s^2\,3d^4$$
Step 2
Actual configuration
One electron shifts from 4s to 3d to achieve a half-filled d-subshell, which has extra stability from symmetrical charge distribution and exchange energy.
$$\text{Actual: } [\text{Ar}]\,4s^1\,3d^5$$
Memorize formulas
Bohr Energy-Emission Relation
Energy difference between two orbits equals the energy of the photon. $h = 6.626 \times 10^{-34}$ J·s.
Formula
$$\Delta E = E_2 - E_1 = h\nu$$
Angular Momentum Quantization
Angular momentum is an integer multiple of $h/2\pi$. $n = 1, 2, 3, \ldots$
Formula
$$mvr = \frac{nh}{2\pi}$$
Bohr Radius (nth orbit)
Radius of nth orbit for hydrogen: $0.529$ Å when $n=1$. For general atom, multiply by $1/Z$.
Formula
$$r_n = 0.529 \, n^2 \text{ Å}$$
Bohr Radius (general atom)
Radius of nth orbit for any hydrogen-like ion with nuclear charge $Ze^+$. $\varepsilon_0 = 8.85 \times 10^{-12}$ C$^2$J$^{-1}$m$^{-1}$.
Formula
$$r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}$$
Energy of nth Orbit (Joules)
Energy per atom in hydrogen. Constant $= 2.178 \times 10^{-18}$ J. For ions, multiply by $Z^2$.
Formula
$$E_n = -\frac{2.178 \times 10^{-18}}{n^2} \text{ J}$$
Energy of nth Orbit (kJ/mol)
Energy per mole in hydrogen. For $\text{He}^+$ ($Z=2$), multiply by $Z^2 = 4$.
Formula
$$E_n = -\frac{1313.31}{n^2} \text{ kJmol}^{-1}$$
Rydberg Formula (Wave Number)
$R_H = 1.09678 \times 10^7$ m$^{-1}$. $n_1$ = lower level, $n_2$ = higher level. For ions, multiply $R_H$ by $Z^2$.
Formula
$$\bar{\nu} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \text{ m}^{-1}$$
Maximum Electrons per Shell
Shell capacity is $2n^2$. Subshell capacity is $2(2\ell+1)$.
Formula
$$\text{Max electrons in shell } n = 2n^2$$
Review
1
What is the radius of the first Bohr orbit of the hydrogen atom?
Ans
0.529 Å
Review
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Apply Bohr radius formula
Use $r_n = 0.529 \, n^2$ Å with $n = 1$
$$r_1 = 0.529 \times 1^2 = 0.529 \text{ Å}$$
2
What is the angular momentum of an electron in the 3rd Bohr orbit of a hydrogen atom?
Ans
$$\frac{3h}{2\pi}$$
Review
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Apply angular momentum quantization
Use $mvr = \frac{nh}{2\pi}$ with $n = 3$
$$mvr = \frac{3h}{2\pi}$$
3
What is the maximum number of electrons that can occupy the shell with principal quantum number $n = 3$?
Ans
18
Review
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Apply max electrons formula
Use $2n^2$ with $n = 3$
$$2n^2 = 2(3)^2 = 18$$
4
Which statement correctly defines the Aufbau principle?
Ans
Electrons fill orbitals in order of increasing energy, starting with the lowest energy orbital
Review
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Identify each principle
Aufbau: fill lowest energy first; Pauli: unique quantum numbers; Hund: singly fill degenerate orbitals with parallel spins
Aufbau principle → electrons fill orbitals in order of increasing energy
5
In which region of the electromagnetic spectrum does the Lyman series of hydrogen appear?
Ans
Ultraviolet region
Review
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Identify the lower energy level
Lyman series: all transitions terminate at $n = 1$
Large energy gap → high frequency → ultraviolet region
6
What is the energy of an electron in the 2nd Bohr orbit of a hydrogen atom in kJ mol⁻¹?
Ans
−328.33 kJ mol⁻¹
Review
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Apply energy formula
Use $E_n = -\frac{1313.31}{n^2}$ kJ mol⁻¹ with $n = 2$
$$E_2 = -\frac{1313.31}{2^2} = -\frac{1313.31}{4} = -328.33 \text{ kJ mol}^{-1}$$
7
Calculate the wave number of the first line of the Balmer series (transition from $n = 3$ to $n = 2$). Given $R_H = 1.097 \times 10^7$ m⁻¹.
Ans
$1.524 \times 10^6$ m⁻¹
Review
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Identify the transition levels
Balmer series first line: $n_1 = 2$ (lower), $n_2 = 3$ (upper)
Lower level $n_1 = 2$, upper level $n_2 = 3$
Apply Rydberg formula
Calculate $\bar{\nu} = R_H \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]$
$$\bar{\nu} = 1.097 \times 10^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = 1.097 \times 10^7 \times \frac{5}{36}$$
Compute final value
Evaluate the expression
$$\bar{\nu} = 1.524 \times 10^6 \text{ m}^{-1}$$
8
Which set of quantum numbers $(n, l, m, s)$ is NOT possible for an electron in an atom?
Ans
$n = 2, \; l = 2, \; m = 0, \; s = +\frac{1}{2}$
Review
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Check the $l$ range for each set
$l$ must satisfy $0 \leq l \leq n - 1$
For $n = 2$: valid $l$ values are 0 and 1; $l = 2$ is invalid
Verify the remaining options
Check all four sets against quantum number constraints
Sets 1, 2, and 4 satisfy all quantum number rules
9
Carbon ($Z = 6$) has the configuration $1s^2 \, 2s^2 \, 2p^2$. According to Hund's rule, how should the two electrons in the 2p subshell be distributed among the three 2p orbitals?
Ans
One electron in each of two orbitals with parallel spins: ↑ ↑ _
Review
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Apply Hund's rule
For 2 electrons in 3 degenerate p orbitals, place one electron in each of two orbitals with parallel spins
2p: ↑ ↑ _ (both electrons unpaired with parallel spins)
10
An electron in a hydrogen atom transitions from $n = 5$ to $n = 2$, emitting a photon. To which spectral series does this transition belong?
Ans
Balmer series
Review
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Identify the lower energy level
The electron drops from $n = 5$ to $n = 2$
Terminal level $n = 2$ → Balmer series
11
What is the ratio of the radius of the 3rd Bohr orbit to the 1st Bohr orbit in a hydrogen atom?
Ans
9:1
Review
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Write the ratio using the formula
$r_3 / r_1 = (0.529 \times 3^2) / (0.529 \times 1^2)$
$$\frac{r_3}{r_1} = \frac{3^2}{1^2} = \frac{9}{1} = 9:1$$
12
An electron in a hydrogen atom has an energy of $-328.33$ kJ mol⁻¹. Which principal quantum number $n$ corresponds to this energy level?
Ans
$n = 2$
Review
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Rearrange the energy formula
From $E_n = -\frac{1313.31}{n^2}$, solve for $n$
$$n^2 = \frac{1313.31}{328.33} = 4.0$$
Solve for $n$
Take the positive square root
$$n = \sqrt{4.0} = 2$$
13
According to the $n + l$ rule, what is the correct order of filling for the orbitals 4s, 3d, and 4p?
Ans
4s → 3d → 4p
Review
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Calculate $n + l$ for each orbital
4s: $4+0=4$, 3d: $3+2=5$, 4p: $4+1=5$
$$4s \rightarrow 3d \rightarrow 4p \quad (\text{lowest } n+l \text{ first; tie broken by lower } n)$$
14
What is the approximate wavelength of the photon emitted when an electron transitions from $n = 4$ to $n = 2$ in a hydrogen atom?
Ans
486 nm
Review
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Calculate energy difference
$\Delta E = 1313.31 \left(\frac{1}{4} - \frac{1}{16}\right)$ kJ mol⁻¹
$$\Delta E = 1313.31 \times \frac{3}{16} = 246.25 \text{ kJ mol}^{-1}$$
Convert to wavelength
$\lambda = \frac{hcN_A}{\Delta E}$
$$\lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)(6.022 \times 10^{23})}{246.25 \times 10^3} \approx 486 \text{ nm}$$
15
Which of the following is the correct ground state electronic configuration of chromium ($Z = 24$)?
Ans
Ar 4s¹ 3d⁵
Review
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Determine expected configuration
Following Aufbau: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
Expected: Ar 4s² 3d⁴
Apply stability exception
A half-filled d⁵ subshell gains extra exchange energy, making Ar 4s¹ 3d⁵ more stable than Ar 4s² 3d⁴
Actual ground state: Ar 4s¹ 3d⁵
Question 1 of 15
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1
What is the radius of the first Bohr orbit of the hydrogen atom?