On this page
The first numericals in Class 11 Structure of Atom look easy: a wavelength goes in, a frequency comes out. Yet many students lose marks here, almost always on units rather than ideas. This post sets out the three relations, works four examples step by step, lists the slips that cost marks, and ends with a short practice routine.
Where does this sit in the syllabus?
Section 2.3.1 of the NCERT Class 11 Chemistry chapter Structure of Atom introduces the wave nature of electromagnetic radiation, the relation c = νλ, and wavenumber, which it defines as "the number of wavelengths per unit length". Its unit is m⁻¹ in SI, though NCERT notes that cm⁻¹ is the unit most commonly used. The JEE (Main) 2026 syllabus opens its Atomic Structure unit with "Nature of electromagnetic radiation", so this is tested in board exams and JEE Main alike. Check the syllabus for your own exam year.
How are wavelength, frequency and wavenumber related?
There are three relations, and the rest follows from them:
c = νλ, where c is the speed of light, ν the frequency and λ the wavelength.
Wavenumber = 1/λ. It is written as nu with a bar over it.
E = hν for the energy of one photon. Combining the three gives E = hc/λ = hc × wavenumber.
NIST lists the speed of light in vacuum as exactly 299 792 458 m s⁻¹, and NCERT rounds it to 3.0 × 10⁸ m s⁻¹, which is what you should use in school problems. NCERT gives Planck's constant h as 6.626 × 10⁻³⁴ J s.
Note the direction of each relation. Frequency, wavenumber and energy all rise together. Wavelength moves the other way. That is why spectroscopists like wavenumber: double the wavenumber and you double the photon energy.
Worked example 1: green light at 500 nm
Find the frequency, wavenumber and photon energy of light with λ = 500 nm.
Convert: λ = 500 nm = 500 × 10⁻⁹ m = 5.00 × 10⁻⁷ m.
Frequency: ν = c/λ = (3.0 × 10⁸) / (5.00 × 10⁻⁷) = 6.0 × 10¹⁴ Hz.
Wavenumber: 1/λ = 1/(5.00 × 10⁻⁷ m) = 2.0 × 10⁶ m⁻¹. In cm⁻¹, divide by 100: 2.0 × 10⁴ cm⁻¹.
Energy per photon: E = hν = 6.626 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J.
Energy per mole of photons: 3.98 × 10⁻¹⁹ × 6.022 × 10²³ = 2.39 × 10⁵ J, or about 239 kJ mol⁻¹.
Worked example 2: from wavenumber back to wavelength
A table lists a band at 1.5 × 10⁴ cm⁻¹ (a sample value for practice). Find λ and ν.
λ = 1/(1.5 × 10⁴ cm⁻¹) = 6.67 × 10⁻⁵ cm = 6.67 × 10⁻⁷ m = 667 nm.
For frequency, keep the units matched. With wavenumber in cm⁻¹, use c = 3.0 × 10¹⁰ cm s⁻¹: ν = 3.0 × 10¹⁰ × 1.5 × 10⁴ = 4.5 × 10¹⁴ Hz.
Is that sensible? NCERT's Problem 2.4 puts the visible range at 400 nm to 750 nm, or 4.0 × 10¹⁴ to 7.5 × 10¹⁴ Hz. So 667 nm is red light, inside the range.
Worked example 3: NCERT's yellow light and radio station
NCERT Problem 2.5 asks for the wavenumber of yellow light at 5800 Å. Since 1 Å = 10⁻¹⁰ m = 10⁻⁸ cm, λ = 5800 × 10⁻⁸ cm, so the wavenumber is 1.724 × 10⁴ cm⁻¹, or 1.724 × 10⁶ m⁻¹. The frequency is 3.0 × 10⁸ / (5800 × 10⁻¹⁰) = 5.17 × 10¹⁴ Hz.
Problem 2.3 uses a radio station broadcasting at 1,368 kHz: λ = 3.0 × 10⁸ / 1.368 × 10⁶ = 219 m, a radio wave. Watch the prefix: kilo is 10³ and mega is 10⁶.
Which unit slips cost the most marks?
cm⁻¹ against m⁻¹. One per centimetre is a hundred per metre, so 1 cm⁻¹ = 100 m⁻¹. Students often divide when they should multiply.
Mixed units in ν = c × wavenumber. Wavenumber in cm⁻¹ needs c in cm s⁻¹ (3.0 × 10¹⁰). Wavenumber in m⁻¹ needs c in m s⁻¹.
Ångström and nanometre. 1 Å = 10⁻¹⁰ m and 1 nm = 10⁻⁹ m. Write the power of ten next to the number before you divide.
Frequency and wavenumber confused. Both are "per something", but frequency is per second and wavenumber is per length.
Per photon or per mole. E = hν gives joules per photon. Multiply by Avogadro's number only when the question asks per mole.
Where does wavenumber show up next?
The Rydberg formula for hydrogen's spectral lines is written in wavenumbers, and NCERT exercise 2.17 asks for the wavenumber of the longest-wavelength Balmer line. Our guide to hydrogen spectrum and Bohr model problems picks up from here. The photon energy E = hν is also the starting point for photoelectric effect problems.
A practice routine for this topic
Write the chain λ, wavenumber, ν, E at the top of the page and mark which way each arrow goes.
Convert every given length to metres before any division. Convert to cm⁻¹ only at the end if the question asks for it.
Solve NCERT Problems 2.3 to 2.5 without looking, then exercises 2.7 and 2.17.
After each answer, run the visible-light check or the radio-wave check.
Next day, mix five questions from this section with two photoelectric questions so you have to pick the relation yourself.
If you still cannot tell where a power of ten went missing, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. Use it to check your working after you have tried each question yourself.
Tonight, convert 400 nm, 500 nm and 750 nm into frequency and wavenumber, both in m⁻¹ and cm⁻¹, and check each against the visible range.
