Atomic StructureHard

Question

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 $10^{4}{\text{ }cm}^{- 1}$. The correct order of energy of these photons is :

Options

A.$C > B > A$
B.$B > A > C$
C.$A > B > C$
D.$A > C > B$

Solution

Wavelength of $A = 400\text{ }nm$.

Wavelength of $B(\lambda) = \frac{C}{v} = \frac{3 \times 10^{8}}{10^{16}} = 3 \times 10^{- 8} = 30 \times 10^{- 9} = 30\text{ }nm$.

Wavelength of $C(\lambda) = \frac{1}{\bar{v}} = \frac{1}{10^{4}} = 10^{- 4}\text{ }cm = 10^{- 6}\text{ }m = 1000\text{ }nm$

Here $\lambda_{C} > \lambda_{A} > \lambda_{B}$

Energy $(E) \propto \frac{1}{\lambda}$

So $E_{B} > E_{A} > E_{C}$.

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