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JEE Mains · Chemistry · STD 11 - 2. structure of atom

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:

  1. A \(C > B > A\)
  2. B \(B > A > C\)
  3. C \(A > B > C\)
  4. D \(A > C > B\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(B > A > C\)

Step-by-step Solution

Detailed explanation

(1) Wavelength of \(A = 400 nm.\) (2) Wavelength of B \((\lambda) = \frac{C}{\nu} = \frac{3 \times 10^{8}}{10^{16}}\) \(=3 \times 10^{-8}=30 \times 10^{-9}=30 nm\). (3) Wavelength of C \((\lambda) = \frac{1}{\bar{\nu}} = \frac{1}{10^{4}} = 10^{-4} cm\) \(=10^{-6} m=1000 nm\)…
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