Question

Consider the following electromagnetic waves: wave A :- wavelength = \(400\,\text{nm}\)
wave B :- frequency = \(10^{16}\,\text{Hz}\)
wave C :- wave number = \(10^{4}\,\text{cm}^{-1}\)
order of energy is :

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

Hint:

For electromagnetic waves, compare {frequencies directly} to decide energy order — higher frequency always means higher photon energy.

Answer 1

The Correct Option is C

Concept:
Energy of an electromagnetic wave (photon) is given by: \[ E = h\nu \] Hence, energy is directly proportional to frequency.

Higher frequency \(\Rightarrow\) higher energy
Wave number \(\bar{\nu}\) is related to wavelength by: \[ \bar{\nu} = \frac{1}{\lambda} \]

Step 1: Find Frequency of Wave A
\[ \lambda_A = 400\,\text{nm} = 4 \times 10^{-7}\,\text{m} \] \[ \nu_A = \frac{c}{\lambda_A} = \frac{3 \times 10^{8}}{4 \times 10^{-7}} = 7.5 \times 10^{14}\,\text{Hz} \]
Step 2: Frequency of Wave B
\[ \nu_B = 10^{16}\,\text{Hz} \]
Step 3: Find Frequency of Wave C
Given wave number: \[ \bar{\nu}_C = 10^{4}\,\text{cm}^{-1} = 10^{6}\,\text{m}^{-1} \] \[ \lambda_C = \frac{1}{\bar{\nu}_C} = 10^{-6}\,\text{m} \] \[ \nu_C = \frac{c}{\lambda_C} = \frac{3 \times 10^{8}}{10^{-6}} = 3 \times 10^{14}\,\text{Hz} \]
Step 4: Compare Energies
\[ \nu_B>\nu_A>\nu_C \] \[ \boxed{B>A>C} \]