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JEE Main Chemistry Structure of Atom 2026 JEE Main 2026 (28 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

The wavelength of photon ' A ' is 400 nm. The frequency of photon ' B ' is 10^ 16 ~s ^ -1 . The wave number of photon ' C^ is 10^ 4 ~cm ^ -1 . The correct order of energy of these photons is :

Options

  1. A. C > B > A
  2. B. ~B > A > C
  3. C. ~A > B > C
  4. D. ~A > C > B

Answer

B. ~B > A > C

Step-by-step solution

Photon A: _A = 400 nm E_A = hc _A = 6.626 10^ -34 3 10^8 400 10^ -9 = 4.97 10^ -19 J Photon B: _B = 10^ 16 s⁻¹ E_B = h _B = 6.626 10^ -34 10^ 16 = 6.626 10^ -18 J Photon C: _C = 10^4 cm⁻¹ = 10^6 m⁻¹ _C = 10^ -6 m E_C = hc _C = 6.626 10^ -34 3 10^8 10^ -6 = 1.988 10^ -19 J Order: E_B > E_A > E_C

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