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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 :

Options

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

Correct 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 = 10¹⁶ s⁻¹ E_B = h _B = 6.626 10⁻³⁴ 10¹⁶ = 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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