Q. The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is 10¹⁶ s⁻¹. The wave number of photon 'C' is 10⁴ cm⁻¹. The correct order of energy of these photons is :

Correct Answer: B > A > C

Solution

Energy of a photon is given by the relation:

E = hν = hc / λ = hc × (wave number)

For photon A:

λ = 400 nm = 4 × 10⁻⁷ m

E_A ∝ 1 / λ = 1 / (4 × 10⁻⁷)

For photon B:

ν = 10¹⁶ s⁻¹

E_B ∝ ν = 10¹⁶

For photon C:

Wave number = 10⁴ cm⁻¹ = 10⁶ m⁻¹

E_C ∝ wave number = 10⁶

Comparing magnitudes:

E_B > E_A > E_C

Hence, the correct order of energy is B > A > C.

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