Start free

Chemistry · Structure of Atom

JEE Main 2026 — 28 January, Evening Shift — Question 57

The wavelength of photon A is 400 nm400\,\mathrm{nm}. The frequency of photon B is 1016 s−110^{16}\,\mathrm{s^{-1}}. The wave number of photon C is 104 cm−110^{4}\,\mathrm{cm^{-1}}.

The correct order of energy of these photons is:

  1. Option A:

    C>B>AC > B > A

  2. Option B:

    B>A>CB > A > C

    Correct
  3. Option C:

    A>B>CA > B > C

  4. Option D:

    A>C>BA > C > B

Answer: B

Step-by-step solution

Energy of photon, E=hν=hcλ=hcνˉE = h\nu = \frac{hc}{\lambda} = hc\bar{\nu}

A: λ=400 nm=4×10−7 m\lambda = 400\,\mathrm{nm} = 4 \times 10^{-7}\,\mathrm{m}

νA=3×1084×10−7=7.5×1014 s−1\nu_A = \frac{3 \times 10^8}{4 \times 10^{-7}} = 7.5 \times 10^{14}\,\mathrm{s^{-1}}

B: νB=1016 s−1\nu_B = 10^{16}\,\mathrm{s^{-1}}

C: νˉ=104 cm−1=106 m−1\bar{\nu} = 10^4\,\mathrm{cm^{-1}} = 10^6\,\mathrm{m^{-1}}

νC=3×108×106=3×1014 s−1\nu_C = 3 \times 10^8 \times 10^6 = 3 \times 10^{14}\,\mathrm{s^{-1}}

On comparison, we have

1016>7.5×1014>3×101410^{16} > 7.5 \times 10^{14} > 3 \times 10^{14}

So, B>A>CB > A > C

Answer key and solution verified before publishing.

Practise Structure of Atom

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Chemistry
Chapter
Structure of Atom
Topic
Electromagnetic Waves and Spectra

More Structure of Atom questions from this paper

The wavelength of photon A is 400\, nm . The frequency of photon B is… | JEE Main 2026 PYQ with Solution · DhiX AI