Atomic Spectra & Nuclear Chemistry

Atomic Spectra & Nuclear Chemistry

Part 1 of 3 — Emission Spectra, Photoelectron Spectroscopy & the Bohr Model

Where this fits. The Atomic Structure & Periodic Trends lesson built the electron model of the atom: quantum numbers, subshells and configurations, and the periodic trends they explain. This lesson tests that model against experiment, first with spectra, the photoelectric effect and photoelectron spectroscopy, then turns to the nucleus itself with radioactive decay and half-life.

These topics appear frequently in MCAT passages because they connect quantum mechanics to experimental data. You won't need to derive the equations — but you must interpret graphs and predict relative values.

The Bohr Model (Hydrogen-Like Atoms)

Bohr proposed that electrons orbit the nucleus only at specific, quantized energy levels.

Energy of Level $n$ in Hydrogen

$E_n = -\frac{13.6\text{ eV}}{n^2} = -\frac{2.18 \times 10^{-18}\text{ J}}{n^2}$

Negative sign: electron is bound to the nucleus (energy $= 0$ at $n = \infty$, ionized).

Ground state: $n=1$, most negative energy, most stable.

Excited state: $n > 1$; higher energy (less negative).

Energy of a Photon Emitted During a Transition

$\Delta E = E_{\text{final}} - E_{\text{initial}} = -13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\text{ eV}$

For a downward transition ($n_i > n_f$): $\Delta E < 0$, photon is emitted.
For an upward transition ($n_f > n_i$): $\Delta E > 0$, photon is absorbed.

$E_{\text{photon}} = h\nu = \frac{hc}{\lambda}$

where $h = 6.626 \times 10^{-34}$ J·s, $c = 3.0 \times 10^8$ m/s.

Spectral Series in Hydrogen

MCAT rule: The largest energy photon in a series is the transition from $n = \infty$ (ionization) to $n_f$. The smallest energy photon is the adjacent transition ($n = n_f + 1$ to $n = n_f$).

Worked Example: Photon Energy

Problem. A hydrogen electron ($E_n = -13.6/n^2$ eV) falls from $n = 3$ to $n = 1$ and emits a photon.

• (a) How much energy does the photon carry, in eV?
• (b) What is that energy in joules?
• (c) What is the photon's wavelength?
• (d) Which series and region of the spectrum is it in?

(a) Energy in eV. The magnitude of the level difference:

$\Delta E = 13.6\left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 13.6 \times \frac{8}{9} = 12.09\text{ eV}$

(b) Energy in joules. With $12.09$ eV from (a): $12.09 \times 1.602 \times 10^{-19} = 1.94 \times 10^{-18}$ J

(c) Wavelength. With $E = 1.94 \times 10^{-18}$ J from (b):

$\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{1.94 \times 10^{-18}} = 1.02 \times 10^{-7}\text{ m} = 102\text{ nm}$

(d) Series and region. The electron ends at $n_f = 1$, so this is the Lyman series; at $102$ nm, shorter than any visible wavelength, it is ultraviolet.

The Photoelectric Effect

• Light shining on a metal ejects electrons only if each photon carries at least the metal's work function $\Phi$, the minimum energy needed to free an electron from the surface.
• The threshold frequency is $f_0 = \Phi / h$. Below it, no electrons leave, however bright the light.
• Above it, the fastest electrons have $KE_{\max} = hf - \Phi$. Higher frequency gives faster electrons; brighter light (more photons) gives MORE electrons, not faster ones.
• This one-photon, one-electron behavior is the evidence that light comes in quanta of energy $hf$.

Photoelectron Spectroscopy (PES)

PES applies the same idea to every subshell of an atom: it bombards a sample with high-energy photons (UV or X-ray), electrons are ejected, and their kinetic energy is measured.

$KE = h\nu - BE$

where $BE$ = binding energy (energy needed to remove the electron from that subshell), the subshell's version of the work function.

Configuration recap: an $s$ subshell holds 2 electrons and a $p$ subshell 6; subshells fill in the order $1s, 2s, 2p, 3s, 3p, 4s$; and $[\text{Ne}]\, 3s^1$ means the 10 electrons of neon ($1s^2\, 2s^2\, 2p^6$) plus one $3s$ electron.

Interpreting a PES Spectrum

• X-axis: Binding energy (increases left to right, or right to left depending on convention — read the axis label!)
• Y-axis: Relative number of electrons (peak area/height)
• Each peak corresponds to one subshell
• Core electrons (inner subshells) have higher binding energy than valence electrons
• Peak relative height ∝ number of electrons in that subshell

Example — Sodium (Na, $[\text{Ne}]\, 3s^1$):

• Peaks at high BE: $1s$ (2 electrons), $2s$ (2 electrons), $2p$ (6 electrons)
• Peak at low BE: $3s$ (1 electron)
• The $3s$ peak is half the height of $2s$ or $1s$ (only 1 electron vs. 2)
• The $2p$ peak is 3× the height of $2s$ (6 electrons vs. 2)

Emission Spectra & Bohr Model 🎯

Wave-Particle Duality & de Broglie Wavelength

Louis de Broglie proposed that matter has wave-like properties. For a particle with momentum $p = mv$:

$\lambda = \frac{h}{mv}$

MCAT implication: Electrons (small mass) have observable wavelengths. Macroscopic objects have negligible wavelengths.

Higher velocity → shorter wavelength (more momentum → smaller de Broglie wave).

Worked Example — An Electron's de Broglie Wavelength

Problem. An electron ($m = 9.1 \times 10^{-31}$ kg) moves at $1.0 \times 10^6$ m/s. A $0.15$ kg baseball moves at $40$ m/s. Use $h = 6.626 \times 10^{-34}$ J·s.

• (a) What is the electron's momentum?
• (b) What is its de Broglie wavelength?
• (c) What is the baseball's wavelength, and why is it never observed?

(a) Momentum. $p = mv = (9.1 \times 10^{-31})(1.0 \times 10^6) = 9.1 \times 10^{-25}$ kg·m/s

(b) Wavelength. With $p$ from (a): $\lambda = \dfrac{h}{p} = \dfrac{6.626 \times 10^{-34}}{9.1 \times 10^{-25}} \approx 7.3 \times 10^{-10}$ m $= 0.73$ nm, comparable to the spacing of atoms in a crystal, so electrons diffract off crystals.

(c) The baseball. $p = (0.15)(40) = 6.0$ kg·m/s, so $\lambda = 6.626 \times 10^{-34} / 6.0 \approx 1.1 \times 10^{-34}$ m, far smaller than any nucleus: no wave behavior can be detected.

Heisenberg Uncertainty Principle

$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$

You cannot simultaneously know both position and momentum exactly. This is why quantum mechanics uses probability orbitals (regions where electrons are likely to be) rather than fixed orbits.

PES Interpretation & Wave-Particle Duality 🎯

Key Takeaways — Part 1

• Bohr transitions. In hydrogen $E_n = -13.6/n^2$ eV; a fall emits a photon of $E = hc/\lambda$, a rise absorbs one.
• Spectral series. Lyman ends at $n = 1$ (UV), Balmer at $n = 2$ (visible), Paschen at $n = 3$ (infrared).
• Reading PES. Peak area counts a subshell's electrons; core electrons sit at the highest binding energy, valence the lowest.
• Matter waves. $\lambda = h/(mv)$ is measurable for electrons, not baseballs; position and momentum can't both be exact.

Common traps

• Sign of $\Delta E$. A downward jump has $\Delta E < 0$ and emits; the photon's energy is the magnitude $|\Delta E|$.
• Judging by the gap in $n$. Levels crowd together at high $n$, so $6 \to 5$ releases far less energy than $2 \to 1$.
• Peak height as energy. In PES, peak size counts electrons; peak position (binding energy) shows how tightly they are held.
• Mixing eV and joules. Convert eV to J ($\times 1.602 \times 10^{-19}$) before using $\lambda = hc/E$ with SI constants.

Atomic Spectra & Nuclear Chemistry

Part 2 of 3 — Nuclear Chemistry: Radioactive Decay & Half-Life

Nuclear chemistry is a distinct but related MCAT topic. You need to recognize decay types, balance nuclear equations, and calculate remaining quantity using half-life.

Types of Radioactive Decay

where $A$ = mass number (protons + neutrons), $Z$ = atomic number (protons).

Alpha decay example: $^{238}_{92}\text{U} \to {}^{234}_{90}\text{Th} + {}^4_2\text{He}$ — $A$ drops by 4 and $Z$ by 2.

Beta-minus decay example: a neutron converts to a proton, $^{14}_{6}\text{C} \to {}^{14}_{7}\text{N} + {}^0_{-1}e + \bar{\nu}$ — $A$ is unchanged and $Z$ rises by 1.

Conservation Laws in Nuclear Reactions

1. Mass number ($A$) is conserved: sum of $A$ on left = sum on right.
2. Atomic number ($Z$) is conserved: sum of $Z$ on left = sum on right.
3. Charge is conserved.

These laws let you identify unknown products.

Worked Example — Balancing a Beta-Minus Decay

Problem. Thorium-234, $^{234}_{90}\text{Th}$, undergoes beta-minus decay, emitting $^{0}_{-1}e$ and an antineutrino.

• (a) What is the mass number of the product?
• (b) What is its atomic number?
• (c) Which nuclide is formed?

(a) Mass number. The beta particle carries no mass number: $234 = A_{\text{product}} + 0$, so $A_{\text{product}} = 234$, unchanged.

(b) Atomic number. $90 = Z_{\text{product}} + (-1)$, so $Z_{\text{product}} = 91$.

(c) The product. With $A = 234$ from (a) and $Z = 91$ from (b), the product is protactinium: $^{234}_{90}\text{Th} \to {}^{234}_{91}\text{Pa} + {}^{0}_{-1}e + \bar{\nu}$.

Half-Life

The half-life $t_{1/2}$ is the time for half of a radioactive sample to decay.

$N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}} = N_0\, e^{-\lambda t}$

where $\lambda = \ln 2 / t_{1/2}$ (this $\lambda$ is the decay constant, not a wavelength).

MCAT shortcut: Work in units of half-lives.

Worked Example — An Iodine-131 Countdown

Problem. A 40 g sample of $^{131}\text{I}$ ($t_{1/2} = 8$ days) is left to decay for 32 days.

• (a) How many half-lives pass?
• (b) What fraction of the sample remains?
• (c) How many grams are left, and how many have decayed?

(a) Half-lives. $32 \div 8 = 4$ half-lives.

(b) Fraction left. With 4 half-lives from (a): $(1/2)^4 = 1/16$, or $6.25\%$.

(c) Mass. $\text{Remaining} = 40 \times 1/16 = \boxed{2.5\text{ g}}$, so $40 - 2.5 = 37.5$ g has decayed.

Binding Energy & Mass Defect

Protons repel one another electrically, yet nuclei hold together. The strong nuclear force does it: it attracts protons and neutrons alike, is far stronger than the electrostatic repulsion, and acts only over about the width of a nucleus (around $10^{-15}$ m).

The nucleus is lighter than the sum of its individual protons and neutrons. This mass defect $\Delta m$ is the mass converted to energy that holds the nucleus together.

$E = \Delta m \cdot c^2 \quad (\text{Einstein's equation})$

With masses in atomic mass units, use $1\text{ u} = 931.5\text{ MeV}$ of energy, so $E\,(\text{MeV}) = \Delta m\,(\text{u}) \times 931.5$.

Binding energy per nucleon is maximized around iron-56 ($^{56}\text{Fe}$) — this is the most stable nucleus. Elements heavier than Fe release energy by fission; lighter elements release energy by fusion.

Worked Example — Binding Energy of Helium-4

Problem. A $^4_2\text{He}$ nucleus has a mass of $4.0015$ u. A proton is $1.0073$ u and a neutron $1.0087$ u.

• (a) What is the mass of the separate nucleons?
• (b) What is the mass defect?
• (c) What is the binding energy, and the binding energy per nucleon?

(a) Separate nucleons. Two protons and two neutrons: $2(1.0073) + 2(1.0087) = 4.0320$ u.

(b) Mass defect. With $4.0320$ u from (a): $\Delta m = 4.0320 - 4.0015 = 0.0305$ u.

(c) Binding energy. $E = 0.0305 \times 931.5 \approx 28.4$ MeV for the nucleus, or $28.4 / 4 \approx 7.1$ MeV per nucleon.

Nuclear Decay & Balancing 🎯

Half-Life Applications in Medicine

Nuclear medicine relies on radioactive isotopes with appropriate half-lives:

• $^{99m}\text{Tc}$ ($t_{1/2} = 6$ h): Most common medical imaging isotope. Short half-life minimizes patient radiation dose. Emits gamma rays, which pass out of the body to an external camera.
• $^{131}\text{I}$ ($t_{1/2} = 8$ d): Treats thyroid cancer and hyperthyroidism. Iodine concentrates in the thyroid gland; beta emission destroys tissue locally.
• $^{18}\text{F}$ ($t_{1/2} = 110$ min): Used in PET scanning (positron emission). Its short half-life means it must be made close to where it is used, within hours.

• The camera that images $^{99m}\text{Tc}$ is a SPECT scanner, and $^{18}\text{F}$ is produced in a cyclotron, usually a regional one that ships doses the same day.

Radiocarbon Dating

$^{14}\text{C}$ ($t_{1/2} = 5{,}730$ years) is used to date organic materials up to ~50,000 years old. Living organisms maintain a constant ratio of $^{14}\text{C}/^{12}\text{C}$ through metabolism. After death, $^{14}\text{C}$ decays. Measuring the remaining ratio gives the age.

Worked Example — Radiocarbon Age of a Wood Sample

Problem. A wood sample contains 12.5% of its original $^{14}\text{C}$ ($t_{1/2} = 5{,}730$ years).

• (a) How many half-lives have passed?
• (b) How old is the sample?
• (c) What percentage will remain after another 5,730 years?

(a) Half-lives. $12.5\% = 1/8 = (1/2)^n$, so $n = 3$ half-lives.

(b) Age. With 3 half-lives from (a): Age $= 3 \times 5{,}730 = \boxed{17{,}190\text{ years}}$

(c) One more half-life. Each half-life removes half of what is LEFT: $12.5\% \div 2 = 6.25\%$ remains, at an age of $4 \times 5{,}730 = 22{,}920$ years.

Half-Life Calculations 🎯

Key Takeaways — Part 2

• Alpha decay. An alpha particle, $^4_2\text{He}$, leaves: $A$ falls by 4 and $Z$ by 2, and both totals balance.
• Beta and capture. $\beta^-$ turns a neutron into a proton ($Z + 1$); $\beta^+$ and electron capture do the reverse ($Z - 1$).
• Count half-lives. $N = N_0(1/2)^{t/t_{1/2}}$: divide the elapsed time by $t_{1/2}$, then halve that many times.
• Binding energy peaks at iron. The mass defect is binding energy, $E = \Delta m c^2$; heavy nuclei fission, light ones fuse.

Common traps

• Changing $A$ in beta decay. $\beta^-$, $\beta^+$ and electron capture all leave $A$ unchanged; only $Z$ shifts, by 1.
• Gamma changing the element. Gamma emission only releases energy: $A$ and $Z$ stay exactly the same.
• Subtracting instead of halving. Each half-life halves what is LEFT, so two half-lives leave 25%, not 0%.
• Long half-lives for imaging. Diagnostic isotopes need SHORT half-lives, like $^{99m}\text{Tc}$ (6 h), to limit the dose.

• Low-yield extras: the SPECT camera used for Tc-99m imaging; cyclotron production of F-18.

Atomic Spectra & Nuclear Chemistry

Part 3 of 3 — Mixed MCAT Review: Passage-Style Practice

This final part uses integrated questions that combine hydrogen spectra, photoelectron spectroscopy, and nuclear chemistry — exactly as the MCAT mixes topics within a single passage.

Quick-Reference Summary

Bohr model (hydrogen): $E_n = -\frac{13.6\text{ eV}}{n^2}$

Half-life: $N = N_0 \left(\frac{1}{2}\right)^{n_{\text{half-lives}}}$

Integrated MCAT-Style Questions — Set 1 🎯

Integrated MCAT-Style Questions — Set 2 🎯

Key Takeaways — Part 3

• Spectra and PES. Photon energy is $13.6\,|1/n_f^2 - 1/n_i^2|$ eV in hydrogen; PES peak area counts electrons.
• Nuclear bookkeeping. Conserve $A$ and $Z$ in every decay, and count half-lives as $(1/2)^n$.

Common traps

• Large $n$, large energy. Levels crowd together at high $n$, so $6 \to 5$ emits far less than $3 \to 2$.
• One step of a chain. Apply each decay to the previous daughter: Ra-226 by alpha then $\beta^-$ ends at $^{222}_{87}\text{Fr}$.

Next up: the Stoichiometry: Moles & Reactions lesson — moles, molar mass and empirical formulas, then limiting reagents, percent yield and reaction types.