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 :

The energy of a photon is given by $$E = h\nu$$, where $$h$$ is Planck’s constant and $$\nu$$ is the frequency. Since $$h$$ is constant, the energy is proportional to the frequency. Therefore, we need to find the frequencies of photons A, B, and C to compare their energies.

We have photon A with wavelength $$\lambda_A = 400 \text{ nm}$$, photon B with frequency $$\nu_B = 10^{16} \text{ s}^{-1}$$, and photon C with wave number $$\bar{\nu}_C = 10^4 \text{ cm}^{-1}$$.

First, we convert all quantities to SI units. For photon A, $$\lambda_A = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4 \times 10^{-7} \text{ m}$$. For photon C, since $$1 \text{ cm} = 0.01 \text{ m}$$, we have $$\bar{\nu}_C = 10^4 \text{ cm}^{-1} = 10^4 \times 100 \text{ m}^{-1} = 10^6 \text{ m}^{-1}$$.

Next, using the speed of light $$c = 3 \times 10^8 \text{ m/s}$$, the frequency relates to wavelength and wave number by $$\nu = \frac{c}{\lambda}$$ and $$\nu = c\,\bar{\nu}$$. For photon A, $$\nu_A = \frac{c}{\lambda_A} = \frac{3 \times 10^8}{4 \times 10^{-7}} = \frac{3}{4} \times 10^{15} = 7.5 \times 10^{14} \text{ Hz}$$. Photon B has $$\nu_B = 10^{16} \text{ Hz}$$ directly. For photon C, $$\nu_C = c \times \bar{\nu}_C = (3 \times 10^8) \times (10^6) = 3 \times 10^{14} \text{ Hz}$$.

The frequencies are $$\nu_A = 7.5 \times 10^{14} \text{ Hz}$$, $$\nu_B = 10^{16} \text{ Hz} = 100 \times 10^{14} \text{ Hz}$$, and $$\nu_C = 3 \times 10^{14} \text{ Hz}$$. Comparing the numerical coefficients gives $$100 > 7.5 > 3$$, so $$\nu_B > \nu_A > \nu_C$$.

Since energy $$E \propto \nu$$, the order of energy is the same as the order of frequency: $$E_B > E_A > E_C$$, giving B > A > C, which corresponds to option A.