Structure of Atom Part 3

Concept:Energy of a photon is inversely proportional to its wavelength: $E \propto \frac{1}{\lambda}$. Smaller wavelength means higher energy.Explanation:Convert all data to the same unit (nanometers) for comparison.Given: Wavelength of photon A: $\lambda_A = 400 \text{ nm}$.Photon B: frequency $\nu = 10^{16} \text{ s}^{-1}$. Wavelength $\lambda = \frac{c}{\nu} = \frac{3 \times 10^8 \text{ m/s}}{10^{16} \text{ s}^{-1}} = 3 \times 10^{-8} \text{ m} = 30 \text{ nm}$.Photon C: wave number $\bar{\nu} = 10^4 \text{ cm}^{-1}$. Wavelength $\lambda = \frac{1}{\bar{\nu}} = \frac{1}{10^4} \text{ cm} = 10^{-4} \text{ cm} = 10^{-6} \text{ m} = 1000 \text{ nm}$.Order of wavelengths: $\lambda_C (1000 \text{ nm}) > \lambda_A (400 \text{ nm}) > \lambda_B (30 \text{ nm})$.Since energy is inversely proportional to wavelength, the order of energy is opposite: $E_B > E_A > E_C$.Shortcut:Directly compare wavelengths – smallest wavelength gives highest energy. Here B has the smallest wavelength (30 nm), then A (400 nm), then C (1000 nm). So energy order is B > A > C.Answer:B $>$ A $>$ C, which corresponds to option C.