Question

The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is \(10^{16}\,\text{s}^{-1}\). The wave number of photon 'C' is \(10^{5}\,\text{cm}^{-1}\). The correct order of energy of these photons is:

• A. C>B>A
• B. B>A>C
• C. A>C>B
• D. A>B>C

Hint:

Always convert wavelength, frequency, and wave number into SI units before calculating photon energy.

Answer 1

The Correct Option is A

Concept: Energy of a photon is given by: \[ E = h\nu = \frac{hc}{\lambda} = hc\bar{\nu} \] where \(\nu\) = frequency, \(\lambda\) = wavelength, \(\bar{\nu}\) = wave number.
Step 1: Energy of photon A. \[ \lambda_A = 400\,\text{nm} = 4 \times 10^{-7}\,\text{m} \] \[ E_A = \frac{hc}{\lambda_A} \approx \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{4 \times 10^{-7}} \approx 4.97 \times 10^{-19}\,\text{J} \]
Step 2: Energy of photon B. \[ \nu_B = 10^{16}\,\text{s}^{-1} \] \[ E_B = h\nu_B = 6.626 \times 10^{-34} \cdot 10^{16} \approx 6.63 \times 10^{-18}\,\text{J} \]
Step 3: Energy of photon C. \[ \bar{\nu}_C = 10^{5}\,\text{cm}^{-1} = 10^{7}\,\text{m}^{-1} \] \[ E_C = hc\bar{\nu}_C = 6.626 \times 10^{-34} \cdot 3 \times 10^{8} \cdot 10^{7} \approx 1.99 \times 10^{-18}\,\text{J} \]
Step 4: Compare energies. \[ E_B (6.63 \times 10^{-18})>E_C (1.99 \times 10^{-18})>E_A (4.97 \times 10^{-19}) \] Thus, the correct order is: \[ C>B>A \]