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:
- A \(C > B > A\)
- B \(B > A > C\)
- C \(A > B > C\)
- D \(A > C > B\)
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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