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

(B) $\mathrm{B}>\mathrm{A}>\mathrm{C}$

(A) $\mathrm{C}>\mathrm{B}>\mathrm{A}$

(D) $\mathrm{A}>\mathrm{C}>\mathrm{B}$

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Atomic Structure

The wavelength of photon '

A

' is 400 nm . The frequency of photon ' B ' is

1 0^{16} s^{- 1}

. The wave number of photon ' C ' is

1 0^{4} cm^{- 1}

. The correct order of energy of these photons is :

Answer: (B)

Photon energy varies inversely with wavelength,

E = \frac{h c}{λ}

, so each photon is reduced to its wavelength. Photon A has

λ = 400 nm

. Photon B has

λ = \frac{c}{ν} = \frac{3 \times 1 0 ^{8}}{1 0 ^{16}} = 3 \times 1 0^{- 8} m = 30 nm

. Photon C has

λ = \frac{1}{ν ˉ} = \frac{1}{1 0 ^{4} c m ^{- 1}} = 1 0^{- 4} cm = 1000 nm

. The wavelengths rank

30 nm < 400 nm < 1000 nm

, so the energies rank in the opposite order,

B > A > C

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option