Question

Consider the following electromagnetic waves A, B and C: (i) The wavelength of A is \(400\,\text{nm}\).
(ii) The frequency of B is \(10^{16}\,\text{s}^{-1}\).
(iii) Wave number of C is \(10^{4}\,\text{cm}^{-1}\).
The correct order of energies is:

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

Hint:

For electromagnetic waves, always compare {frequencies} to decide energy order. Higher frequency $\Rightarrow$ higher photon energy.

Answer 1

The Correct Option is B

Concept:
Energy of an electromagnetic wave (photon) is directly proportional to its frequency: \[ E = h\nu \] Hence, higher frequency implies higher energy.
Step 1: Find Frequency of Wave A
\[ \lambda_A = 400\,\text{nm} = 4 \times 10^{-7}\,\text{m} \] \[ \nu_A = \frac{c}{\lambda_A} = \frac{3 \times 10^{8}}{4 \times 10^{-7}} = 7.5 \times 10^{14}\,\text{s}^{-1} \]
Step 2: Frequency of Wave B
\[ \nu_B = 10^{16}\,\text{s}^{-1} \]
Step 3: Find Frequency of Wave C
Given wave number: \[ \bar{\nu}_C = 10^{4}\,\text{cm}^{-1} = 10^{6}\,\text{m}^{-1} \] \[ \lambda_C = \frac{1}{\bar{\nu}_C} = 10^{-6}\,\text{m} \] \[ \nu_C = \frac{c}{\lambda_C} = \frac{3 \times 10^{8}}{10^{-6}} = 3 \times 10^{14}\,\text{s}^{-1} \]
Step 4: Compare Energies
\[ \nu_B>\nu_A>\nu_C \] \[ \boxed{B>A>C} \]