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MHT-CET Chemistry · Structure of Atom

Electromagnetic Radiation and Wave Properties

Light is an electromagnetic wave described by its wavelength, frequency and wavenumber; all such waves travel at the speed of light (c = nu*lambda), carry energy in quantised photons (E = h*nu = hc/lambda), and line up in a fixed spectrum from low-energy radio waves to high-energy gamma rays.

Why this matters

This is one of the most reliable scoring blocks in MHT-CET Chemistry Structure of Atom — almost every PYQ is a one-step plug-in: find a frequency from a wavelength, a wavelength from a wavenumber, or a photon energy from Planck's relation. The rest test a single recalled fact: which colour or radiation has the highest or lowest energy. Learn c = nu*lambda, E = hc/lambda and the spectrum order (radio to gamma) cold, keep every quantity in SI, and you can attempt every question here on sight.

Concept 1 of 4: Wave characteristics — wavelength, frequency, wavenumber, amplitude

A travelling wave is described by four numbers. Wavelength is the length of one full cycle, frequency is how many cycles pass a fixed point each second, wavenumber is simply how many wavelengths fit in one metre (or cm), and amplitude is the height of the crest. The bank's favourite one-liner is 'the number of waves passing a point per second' — that is frequency, not wavelength.

Definition

The four wave parameters:

  • Wavelength λ\lambda — the distance of one complete wave (crest to crest); measured in metres, nm or angstrom.
  • Frequency ν\nu — the number of waves passing a given point per second; unit hertz (Hz=s−1\text{Hz} = \text{s}^{-1}).
  • Wavenumber νˉ\bar{\nu} — the number of waves per unit length, νˉ=1λ\bar{\nu} = \dfrac{1}{\lambda}; unit m−1\text{m}^{-1} or cm−1\text{cm}^{-1}.
  • Amplitude — the maximum displacement (height) of the wave; it sets the brightness/intensity, NOT the energy of a photon.
  • The wave travels at velocity cc, linking these by c=νλc = \nu\lambda.

Wavenumber

νˉ=1λ\bar{\nu} = \dfrac{1}{\lambda}
  • νˉ\bar{\nu}wavenumber (m^-1 or cm^-1)
  • λ\lambdawavelength (m or cm, matching the wavenumber unit)

Worked example

What is the wavenumber (in m^-1) of a radiation whose wavelength is 500 nm?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q62Easy

Example 1 · Structure of Atom · Electromagnetic Radiation and Wave Properties

What is wavenumber of a radiation having wavelength 0.25 μm0.25\,\mu\text{m}?

Frequency vs wavelength — read the wording

'Number of waves passing a point per second' is frequency ν\nu. Wavelength is the length of ONE wave (a distance), and wavenumber νˉ=1/λ\bar{\nu} = 1/\lambda is waves per unit LENGTH — do not confuse the two 'number of waves' phrasings.

Match the wavenumber unit to lambda

νˉ=1/λ\bar{\nu} = 1/\lambda gives cm−1\text{cm}^{-1} only if λ\lambda is in cm, and m−1\text{m}^{-1} only if λ\lambda is in metres. Convert μm\mu\text{m} or nm to the required base unit FIRST (0.25 μm=0.25×10−6 m0.25\ \mu\text{m} = 0.25 \times 10^{-6}\ \text{m}).

Concept 2 of 4: Speed of light relation, c = nu*lambda

Every electromagnetic wave — radio, light, X-ray — travels through vacuum at the same speed c=3×108 m/sc = 3 \times 10^{8}\ \text{m/s}. Because that speed is fixed, wavelength and frequency are locked in an inverse trade-off: a shorter wavelength must have a higher frequency. This one relation answers almost every 'find the frequency from the wavelength' PYQ in a single step.

Definition

The velocity relation for electromagnetic radiation:

  • c=νλc = \nu\lambda, so ν=cλ\nu = \dfrac{c}{\lambda} and λ=cν\lambda = \dfrac{c}{\nu}.
  • c=3×108 m/sc = 3 \times 10^{8}\ \text{m/s} is the same for ALL electromagnetic radiation in vacuum.
  • At fixed cc, frequency and wavelength are inversely proportional: ν∝1λ\nu \propto \dfrac{1}{\lambda}.
  • Keep λ\lambda in metres so that ν\nu comes out in Hz (convert nm by ×10−9\times 10^{-9}).

Speed of light relation

c=νλ⇒ν=cλc = \nu\lambda \qquad \Rightarrow \qquad \nu = \dfrac{c}{\lambda}
  • cspeed of light, 3 x 10^8 m/s (same for all EM radiation)
  • ν\nufrequency (Hz)
  • λ\lambdawavelength (m)

Worked example

Calculate the frequency of light of wavelength 600 nm. (c = 3 x 10^8 m/s)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q74Easy

Example 2 · Structure of Atom · Electromagnetic Radiation and Wave Properties

What is the frequency of violet light having wavelength 400 nm?

Convert nm to metres before dividing

For ν=c/λ\nu = c/\lambda in Hz, λ\lambda must be in metres. Forgetting the ×10−9\times 10^{-9} on a nm wavelength shifts the answer by nine orders of magnitude. 400 nm=400×10−9 m=4×10−7 m400\ \text{nm} = 400 \times 10^{-9}\ \text{m} = 4 \times 10^{-7}\ \text{m}.

c is the same for every EM radiation

Radio waves, visible light and gamma rays all travel at 3×108 m/s3 \times 10^{8}\ \text{m/s} in vacuum. What differs between them is λ\lambda and ν\nu, never cc.

Concept 3 of 4: Planck's quantum theory and photon energy, E = h*nu = hc/lambda

Planck showed that energy is not radiated continuously but in tiny discrete packets called quanta (a quantum of light is a photon). The energy of one photon is set purely by its frequency — higher frequency means a more energetic photon. Rewriting frequency as c/λc/\lambda gives the form the bank loves: E=hc/λE = hc/\lambda, so a shorter-wavelength photon carries MORE energy.

Definition

Planck's quantum theory:

  • Radiant energy is emitted or absorbed only in whole-number multiples of a quantum: E=nhνE = nh\nu (n = 1, 2, 3, ...).
  • The energy of ONE photon is E=hν=hcλE = h\nu = \dfrac{hc}{\lambda}.
  • Energy is directly proportional to frequency and inversely proportional to wavelength: high ν\nu / short λ\lambda = high energy.
  • Planck's constant h=6.626×10−34 J sh = 6.626 \times 10^{-34}\ \text{J s}.
  • Energy per mole of photons =E×NA= E \times N_A, where NA=6.022×1023N_A = 6.022 \times 10^{23}.

Photon energy (Planck)

E=hν=hcλ(per mole: E×NA)E = h\nu = \dfrac{hc}{\lambda} \qquad (\text{per mole: } E \times N_A)
  • Eenergy of one photon (J)
  • hPlanck's constant, 6.626 x 10^-34 J s
  • ν\nufrequency (Hz)
  • cspeed of light, 3 x 10^8 m/s
  • λ\lambdawavelength (m)
  • N_AAvogadro number, 6.022 x 10^23 mol^-1

Worked example

Calculate the energy of one photon of light of wavelength 400 nm. (h = 6.626 x 10^-34 J s, c = 3 x 10^8 m/s)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q75Moderate

Example 3 · Structure of Atom · Electromagnetic Radiation and Wave Properties

Calculate the energy per mole of photon of electromagnetic radiation having wavelength 700 nm[ h=6.626×10−34Js,c=3×108 m/s]700\text{ }nm\left\lbrack \text{ }h = 6.626 \times10^{- 34}Js,c = 3 \times10^{8}\text{ }m/s \right\rbrack

Energy goes as 1/lambda, not lambda

Because E=hc/λE = hc/\lambda, a shorter wavelength means a larger energy. Do not assume the longest-wavelength radiation is the most energetic — it is the least energetic.

Per photon vs per mole

E=hc/λE = hc/\lambda gives the energy of a SINGLE photon (∼10−19 J\sim 10^{-19}\ \text{J}). If the question asks 'per mole', you must multiply by NA=6.022×1023N_A = 6.022 \times 10^{23} to reach the ∼105 J\sim 10^{5}\ \text{J} range.

Amplitude does not set energy

A photon's energy depends only on frequency/wavelength, not on amplitude. Amplitude controls intensity (number of photons / brightness), which is a different quantity.

Concept 4 of 4: The electromagnetic spectrum — order by frequency and energy

The electromagnetic spectrum lays out every kind of radiation in one line. Radio waves sit at the low-energy end (longest wavelength, lowest frequency) and gamma rays at the high-energy end (shortest wavelength, highest frequency), with visible light a thin band in the middle. The bank rarely asks for numbers — it asks 'which has the highest/lowest energy', which is a pure recall of this order.

Definition

The spectrum in order of increasing frequency and energy (decreasing wavelength):

  • Radio waves — longest λ\lambda, lowest ν\nu, lowest energy.
  • Microwaves — next up.
  • Infrared (IR) — felt as heat.
  • Visible light — the only band we see; within it the order is VIBGYOR, red lowest energy, violet highest energy.
  • Ultraviolet (UV) — higher energy than visible.
  • X-rays — high energy, penetrating.
  • Gamma rays — shortest λ\lambda, highest ν\nu, highest energy.
Radiation (low to high energy)Wavelength / frequencyEnergy
Radio wavesLongest wavelength, lowest frequencyLowest energy
MHT-CET — of radio waves, microwaves, IR and UV, radio waves have the LOWEST energy.
MicrowavesLong wavelength, low frequencyVery low
Infrared (IR)Longer than visibleLow (felt as heat)
Visible light (VIBGYOR)400–700 nm; red longest, violet shortestRed lowest, violet highest
Within visible light, VIOLET has the highest energy and RED the lowest (energy increases R->V).
Ultraviolet (UV)Shorter than visibleHigher than visible
X-raysVery short wavelengthHigh, penetrating
Gamma raysShortest wavelength, highest frequencyHighest energy
Energy increases from radio waves to gamma rays: E proportional to frequency proportional to 1/wavelength.
Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q57Easy

Example 4 · Structure of Atom · Electromagnetic Radiation and Wave Properties

Which from following coloured light has highest energy?

Long wavelength = LOW energy

Radio waves have the LONGEST wavelength, so by E=hc/λE = hc/\lambda they carry the least energy — a common trap is to pick them as 'highest'. Highest energy always goes to the shortest-wavelength radiation (gamma rays; violet among the visible colours).

VIBGYOR direction

Reading VIBGYOR, violet is at the high-frequency (high-energy) end and red at the low-energy end. Energy rises from Red to Violet, so 'highest energy colour' is violet, not red.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Reference tables (1)

The electromagnetic spectrum — order by frequency and energy7 rows
Radiation (low to high energy)Wavelength / frequencyEnergy
Radio wavesLongest wavelength, lowest frequencyLowest energy
MHT-CET — of radio waves, microwaves, IR and UV, radio waves have the LOWEST energy.
MicrowavesLong wavelength, low frequencyVery low
Infrared (IR)Longer than visibleLow (felt as heat)
Visible light (VIBGYOR)400–700 nm; red longest, violet shortestRed lowest, violet highest
Within visible light, VIOLET has the highest energy and RED the lowest (energy increases R->V).
Ultraviolet (UV)Shorter than visibleHigher than visible
X-raysVery short wavelengthHigh, penetrating
Gamma raysShortest wavelength, highest frequencyHighest energy
Energy increases from radio waves to gamma rays: E proportional to frequency proportional to 1/wavelength.

Watch out for (9)

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