Mark out the correct statement with respect to thermal radiations emitted by a black body

Options

A. At a given temperature, , energy is distributed non - uniformly among different wavelengths
B. As temperature of body is increased , energy content of all wavelengths decreases
C. The product of `E_lamda` (spectral energy) with `D_(lamda)` (spectral width) , for all equal `Deltalamda'` is the same
D. The thermal radiation emitted by a hot body is a discrete spectra

Answer

To solve the question regarding the correct statement with respect to thermal radiations emitted by a black body, we will analyze the properties of black body radiation and the statements provided.

### Step-by-Step Solution:

1. **Understanding Black Body Radiation**:
- A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It also emits radiation in a characteristic spectrum that depends solely on its temperature.

2. **Graph of Black Body Radiation**:
- The intensity of radiation emitted by a black body as a function of wavelength is represented by a continuous spectrum. The graph is typically plotted with intensity (or emissive power) on the y-axis and wavelength on the x-axis. This graph is not symmetrical and shows that energy distribution among different wavelengths is non-uniform.

3. **Analyzing the Statements**:
- **Statement A**: "At a given temperature, energy is distributed non-uniformly among the different wavelengths."
- This statement is correct because the graph of black body radiation shows that the intensity varies with wavelength, indicating a non-uniform distribution.

- **Statement B**: "The temperature of the body increases, the energy content of all wavelengths increases."
- This statement is misleading. While the total energy emitted increases with temperature (according to Stefan-Boltzmann Law), the distribution of energy among wavelengths changes as well. Therefore, this statement is not entirely correct.

- **Statement C**: "The product of \( E_\lambda \) with \( d\lambda \) for \( r = \eta' E_\lambda \) will not be the same."
- This statement is incorrect because the emissive power \( E_\lambda \) multiplied by \( d\lambda \) gives the intensity of radiation at that wavelength, which is a valid expression in the context of black body radiation.

- **Statement D**: "Thermal radiation emitted by a hot body is a discrete spectrum."
- This statement is incorrect. The thermal radiation emitted by a black body is continuous, not discrete.

4. **Conclusion**:
- The only correct statement is **Statement A**.

### Final Answer:
The correct statement with respect to thermal radiations emitted by a black body is: **At a given temperature, energy is distributed non-uniformly among the different wavelengths.**