Q4 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry

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

  • A
    $\mathrm{C}>\mathrm{B}>\mathrm{A}$
  • Correct answer
    B
    $\mathrm{~B}>\mathrm{A}>\mathrm{C}$
  • C
    $\mathrm{~A}>\mathrm{B}>\mathrm{C}$
  • D
    $\mathrm{~A}>\mathrm{C}>\mathrm{B}$

Solution

Photon A: $\lambda_A = 400$ nm
$E_A = \frac{hc}{\lambda_A} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19}$ J
Photon B: $\nu_B = 10^{16}$ s⁻¹
$E_B = h\nu_B = 6.626 \times 10^{-34} \times 10^{16} = 6.626 \times 10^{-18}$ J
Photon C: $\bar{\nu}_C = 10^4$ cm⁻¹ = $10^6$ m⁻¹
$\lambda_C = 10^{-6}$ m
$E_C = \frac{hc}{\lambda_C} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10^{-6}} = 1.988 \times 10^{-19}$ J
Order: $E_B > E_A > E_C$

Hence, the correct answer is option: 'B'.

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