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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.

All Structure of Atom notesMHT-CET Chemistry strategy

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

Intuition

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}
  • \bar{\nu}wavenumber (m^-1 or cm^-1)
  • \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?
  1. Convert the wavelength to metres: λ=500 nm=500×10−9 m=5×10−7 m\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5 \times 10^{-7}\ \text{m}.
  2. Wavenumber is the reciprocal: νˉ=1λ=15×10−7\bar{\nu} = \dfrac{1}{\lambda} = \dfrac{1}{5 \times 10^{-7}}.
Answer:νˉ=2×106 m−1\bar{\nu} = 2 \times 10^{6}\ \text{m}^{-1}.
Practice this conceptself-check · 4 quick reps

Try it yourself

Calculate the wavelength (in nm) of a photon whose wavenumber is 1516 cm−1\dfrac{15}{16}\ \text{cm}^{-1}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Which parameter is the number of waves passing a given point in one second?
  2. 2.
    Wavenumber of a radiation of wavelength 0.25 μm0.25\ \mu\text{m}?
  3. 3.
    SI unit of frequency?
  4. 4.
    Which wave property decides the intensity (brightness) but not the photon energy?

From the bank · past-year question

Example 1Structure of AtomEASY
What is wavenumber of a radiation having wavelength 0.25 μm0.25\,\mu\text{m}?

[Q62 · 16th May Shift 1 · 2023]

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}).
Drill 2 more on wave characteristics — wavelength, frequency, wavenumber, amplitude

Concept 2 of 4

Speed of light relation, c = nu*lambda

Intuition

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)
  • \nufrequency (Hz)
  • \lambdawavelength (m)

Worked example

Calculate the frequency of light of wavelength 600 nm. (c = 3 x 10^8 m/s)
  1. Convert to metres: λ=600 nm=600×10−9 m=6×10−7 m\lambda = 600\ \text{nm} = 600 \times 10^{-9}\ \text{m} = 6 \times 10^{-7}\ \text{m}.
  2. Use ν=cλ=3×1086×10−7\nu = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8}}{6 \times 10^{-7}}.
Answer:ν=5×1014 Hz\nu = 5 \times 10^{14}\ \text{Hz}.
Practice this conceptself-check · 4 quick reps

Try it yourself

Calculate the frequency of blue light having wavelength 440 nm. (c = 3 x 10^8 m/s)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Frequency of radiation of wavelength 750 nm?
  2. 2.
    Frequency of violet light of wavelength 400 nm?
  3. 3.
    What is the speed of X-rays in vacuum?
  4. 4.
    If wavelength doubles at fixed c, what happens to frequency?

From the bank · past-year question

Example 2Structure of AtomEASY
What is the frequency of violet light having wavelength 400 nm?

[Q74 · 10th May Shift 2 · 2023]

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.
Drill 2 more on speed of light relation, c = nu*lambda

Concept 3 of 4

Planck's quantum theory and photon energy, E = h*nu = hc/lambda

Intuition

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
  • \nufrequency (Hz)
  • cspeed of light, 3 x 10^8 m/s
  • \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)
  1. Convert: λ=400 nm=400×10−9 m\lambda = 400\ \text{nm} = 400 \times 10^{-9}\ \text{m}.
  2. Use E=hcλ=(6.626×10−34)(3×108)400×10−9E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{400 \times 10^{-9}}.
  3. E=1.9878×10−254×10−7E = \dfrac{1.9878 \times 10^{-25}}{4 \times 10^{-7}}.
Answer:E≈4.97×10−19 JE \approx 4.97 \times 10^{-19}\ \text{J} per photon.
Practice this conceptself-check · 4 quick reps

Try it yourself

Calculate the energy per mole of photons of radiation of wavelength 500 nm. (h = 6.626 x 10^-34 J s, c = 3 x 10^8 m/s, N_A = 6.022 x 10^23)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Write the two equivalent forms of a photon's energy.
  2. 2.
    Value of Planck's constant (with units)?
  3. 3.
    Photon energy is directly proportional to which quantity?
  4. 4.
    How do you convert one photon's energy to energy per mole?

From the bank · past-year question

Example 3Structure of AtomMODERATE
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

[Q75 · 21 April Shift I · 2025]

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

Intuition

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

Try it yourself

Which of these electromagnetic radiations possesses the lowest energy: radio waves, microwaves, infrared, or ultraviolet?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Which coloured visible light has the highest energy?
  2. 2.
    Which visible colour has the lowest energy?
  3. 3.
    Which EM radiation has the highest energy overall?
  4. 4.
    Which has lower energy: microwaves or ultraviolet?
  5. 5.
    Order red, blue and violet from lowest to highest energy.

From the bank · past-year question

Example 4Structure of AtomEASY
Which from following coloured light has highest energy?

[Q57 · 12th May Shift 2 · 2024]

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.
Drill 1 more on the electromagnetic spectrum — order by frequency and energy

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)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Structure of AtomEASY
Calculate wavelength for emission of a photon having wave number 1516 cm−1\frac{15}{16}\text{ cm}^{-1}.

[Q51 · 4th May Shift 1 · 2023]

Example 2Structure of AtomEASY
Calculate the frequency of blue light having wavelength 440 nm.

[Q85 · 12th May Shift 1 · 2024]

Example 3Structure of AtomEASY
Which of the following electromagnetic radiations possesses lowest energy?

[Q81 · 13th May Shift 2 · 2024]

Example 4Structure of AtomEASY
Which parameter is indicated by the number of waves passing through a given point in one second?

[Q89 · 9th May Shift 2 · 2023]

Example 5Structure of AtomEASY
Calculate the frequency if wavelength is 750 nm.

[Q52 · 14th May Shift 2 · 2024]

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