Step 1: Understanding the Concept:
Electromagnetic radiation propagates as waves with distinct wavelengths ($\lambda$) and frequencies ($\nu$).
The energy ($E$) of a single photon of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength.
Step 2: Key Formula or Approach:
The relationship between photon energy and wavelength is given by Planck's equation:
\[ E = h\nu = \frac{hc}{\lambda} \]
where $h$ is Planck's constant and $c$ is the speed of light.
According to this relationship, shorter wavelengths correspond to higher energy levels.
Step 3: Detailed Explanation:
Let us compare the wavelengths and energy levels of the given electromagnetic regions:
1. Gamma rays (B): These have the shortest wavelengths ($< 10^{-11} \text{ m}$) and the highest frequencies in the electromagnetic spectrum, making them the most energetic radiation.
2. Visible rays (C): These lie in a narrow band with wavelengths ranging from approximately $380 \text{ nm}$ (violet) to $750 \text{ nm}$ (red). Their energy is significantly lower than gamma rays but higher than infrared.
3. Infrared rays (A): These have longer wavelengths ($750 \text{ nm}$ to $1 \text{ mm}$) than visible light, corresponding to lower frequency and lower photon energy.
4. Radio waves (D): These have the longest wavelengths ($1 \text{ mm}$ to $> 100 \text{ km}$) and the lowest frequencies, making them the least energetic radiation in the spectrum.
Arranging these in order of decreasing energy level yields:
Gamma rays (B) $\rightarrow$ Visible rays (C) $\rightarrow$ Infrared rays (A) $\rightarrow$ Radio waves (D).
This corresponds to the sequence: (B), (C), (A), (D).
Step 4: Final Answer:
The correct decreasing energy sequence is (B), (C), (A), (D), corresponding to option (B).