Electromagnetic spectrum recap — UV, visible, IR, NMR, microwave, X-ray

WHY do we care?

Spectroscopy is how chemists see molecules without a microscope. Every region of the EM spectrum has just the right energy to excite one specific event inside matter:

Region	Energy	What it excites
X-ray	highest	knocks out core (inner-shell) electrons
UV–Visible	high	valence electron transitions (π→π, n→π)
Infrared (IR)	medium	bond vibrations (stretch, bend)
Microwave	low	molecular rotations
Radio (NMR)	lowest	nuclear spin flips in a magnetic field

The three master equations (derived, not dumped)

WHY $c=\lambda\nu$ ? Speed = distance ÷ time. In one second, $\nu$ crests pass, and each crest is $\lambda$ metres apart, so the wave front travels $\lambda \times \nu$ metres per second.

$\boxed{c = \lambda\nu}$

HOW energy enters (Planck/Einstein): Light comes in packets (photons). Experiment (photoelectric effect) showed each packet's energy is proportional to frequency, with proportionality constant $h$ (Planck's constant, $6.626\times10^{-34}\ \text{J s}$ ):

$\boxed{E = h\nu}$

Combine the two — substitute $\nu = c/\lambda$ :

$E = h\nu = \frac{hc}{\lambda}$

Figure — Electromagnetic spectrum recap — UV, visible, IR, NMR, microwave, X-ray

Approximate landmarks (memorise these!)

Worked examples

Worked example 1. Energy of a UV photon at

$\lambda = 250\ \text{nm}$ $E=\frac{hc}{\lambda}=\frac{(6.626\times10^{-34})(3.00\times10^8)}{250\times10^{-9}}$

Why convert nm→m? SI: $h$ is in J·s, $c$ in m/s, so $\lambda$ MUST be in metres or units don't cancel. $E=\frac{1.988\times10^{-25}}{2.50\times10^{-7}}=7.95\times10^{-19}\ \text{J}$ Per mole: $\times N_A = 7.95\times10^{-19}\times6.022\times10^{23} \approx 479\ \text{kJ mol}^{-1}$ . Why × $N_A$ ? To compare with bond energies (kJ/mol). 479 kJ/mol can break many bonds — that's why UV light causes photochemistry/sunburn.

Steel-man your mistakes

Active recall

Recall Click to test yourself

Q: Which region has the highest energy photons? → X-ray.
Q: Which molecular event does IR probe? → Bond vibrations.
Q: Write $E$ in terms of $\lambda$ . → $E=hc/\lambda$ .
Q: Why does NMR use radio waves? → Nuclear spin flips need very little energy.
Q: Wavenumber units and definition? → cm⁻¹, $\bar\nu=1/\lambda$ .

Recall Feynman: explain to a 12-year-old

Imagine a row of friends standing close together (short wave) vs spread far apart (long wave). The close-together ones are like an energetic, fast-buzzing line — that's X-ray/UV light, strong enough to knock things over (electrons). The far-apart, lazy ones are radio waves — so gentle they can only make a tiny compass needle inside an atom (the nucleus) wiggle. We shine different kinds of light at a molecule and watch which one it "eats." Whatever it eats tells us what part of it moved — a spinning part, a wobbling spring, or a jumping electron. That's how we figure out what the molecule is made of without ever touching it.

Flashcards

Relationship between c, λ, ν

c=\lambda\nu

(speed = wavelength × frequency)

Energy of a photon in terms of frequency

E=h\nu

h=6.626\times10^{-34}

J·s

Energy of a photon in terms of wavelength

E=hc/\lambda

Is energy proportional or inversely proportional to wavelength?

Inversely:

E\propto 1/\lambda

Order of EM regions by increasing energy

Radio(NMR) < Microwave < IR < Visible < UV < X-ray

Molecular event probed by UV–Visible

Valence electron transitions (π→π*, n→π*)

Molecular event probed by IR

Bond vibrations (stretching, bending)

Molecular event probed by microwave

Molecular rotations

Molecular event probed by NMR

Nuclear spin flips in a magnetic field

Molecular event probed by X-ray

Inner-shell (core) electron transitions / diffraction

Definition of wavenumber

\bar\nu = 1/\lambda

, units cm⁻¹;

E\propto\bar\nu

Visible light wavelength range

~400 nm (violet) to ~700 nm (red)

Why must λ be in metres in E=hc/λ

To keep SI units consistent with h (J·s) and c (m/s)

Convert wavenumber to frequency

\nu = c\,\bar\nu

Connections

UV-Visible Spectroscopy — π→π* and n→π* transitions, Beer–Lambert law
IR Spectroscopy — functional-group fingerprint region, $\bar\nu$ in cm⁻¹
NMR Spectroscopy — chemical shift, spin states in $B_0$
Planck's Quantum Theory — origin of $E=h\nu$
Photoelectric Effect — experimental basis for photon energy
Atomic Structure — Electron Transitions — energy levels and absorption
Beer-Lambert Law — quantitative absorbance

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, light asal mein energy hai jo wave ki tarah travel karti hai. Har "type" ki light ka apna energy level hota hai. Yeh teen formule yaad rakho: $c=\lambda\nu$ (speed = wavelength × frequency), aur $E=h\nu=hc/\lambda$ . Sabse important baat — energy aur frequency dost hain (saath badhte hain), lekin wavelength dushman hai (wavelength badhi to energy ghati). X-ray ka wavelength sabse chhota, isliye energy sabse zyada; radio waves (NMR) ka wavelength sabse bada, energy sabse kam.

Spectroscopy ka funda simple hai: alag-alag light molecule ke alag-alag hisson ko hilati hai. Radio waves (NMR) sirf nucleus ka spin flip karti hain — bahut hi kam energy chahiye. Microwave molecule ko ghumati (rotation) hai. IR bonds ko spring ki tarah stretch/bend karti hai. UV-Visible valence electrons ko upar uchhalti hai. X-ray to andar wale (core) electrons ko hi nikaal deti hai. Energy ladder par jitna upar jaoge, utni badi cheez ko disturb karoge.

Common galti: students sochte hain bada wavelength matlab zyada energy — galat! $\lambda$ denominator mein hai $E=hc/\lambda$ mein, isliye lamba wave = kamzor photon. Doosri galti: numerical mein nm ko metre mein convert karna bhool jaate hain. Hamesha SI units mein lao, warna answer $10^9$ guna galat aata hai. Aur wavenumber ( $\bar\nu=1/\lambda$ , cm⁻¹) ko frequency (Hz) se mat confuse karo — dono alag hain, factor $c$ ka difference hai.

Yaad rakhne ka tareeka: "Radio Mics In Very eXpensive Gear" — Radio < Microwave < IR < Visible < UV < X-ray, energy badhte order mein. Isse exam mein region order kabhi nahi bhoologe.

Test yourself — Spectroscopy & Analysis (Intro)

Connections