Test: Blackbody Radiation - EmSAT Achieve MCQ

The laws of blackbody radiation governed by

Planck's law of black-body radiation: Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T, when there is no net flow of matter and energy between the body and its environment.

Wien's displacement law: Wien's displacement law states that the blackbody radiation curve for different temperature peaks at a wavelength that is inversely proportional to the temperature.

Stefan–Boltzmann law: The Stefan–Boltzmann law describes the power radiated per unit time from a black body in terms of its temperature.

Given,

Power, P = 100 kW

Time, t = 1 hour

∴ E = P × t = 100 × 1 = 100 kWh

E = 100 × 3.6 × 10⁶ J = 36 × 10^{7 J     (∵ 1 kWh = 3.6 × 106 J)}

Hence, option (2) is the correct answer.

Gamma Rays:

• A gamma ray, also known as gamma radiation.
• It is a penetrating form of electromagnetic radiation arising from the radioactive decay of atomic nuclei.
• It consists of the shortest wavelength electromagnetic waves, typically shorter than those of X-rays. 

• They are produced by the hottest and most energetic objects in the universe.

Therefore, radiation with maximum penetration power is gamma rays.

Given that:

The heat radiated at 227∘C, E₁ = 7 cal / cm² / s

Initial temperature of the body, T₁ = 227∘C ⇒ (227 + 273) K = 500 K

Final temperature of the body, T₂ =  727∘C ⇒ (727 + 273) K = 1000 K

 

• A black body is an example of a good absorber of heat as well as a good emitter of heat.
• The ease with which a black body can absorb a photon is the reverse process of emitting the one.
• This entire cycle takes place because of the number of transitions that are associated with the EM field.

The Energy Distribution curve as shown in the figure concludes that:

• As the wavelength increases, the energy emitted increases and reaches the maximum height. After this point, it started decreasing gradually.
• Hence we can say that the Energy distribution of black body radiation state that when wavelength increases, the energy emitted from the black body radiation will increase first and then decreases.
• Hence option 2 is correct.

• A black body is an idealized physical body that absorbs all incident electromagnetic radiation of any wavelengths, regardless of frequency or angle of incidence. So option 4 is correct.

• When the body will be in black colors all the light gets absorbed because black is the good absorber where white light emits or reflect all the light coming towards it.
• Application of White colors:
  • Radiator in homes is painting with white.
  • We wear white cloth in summer because it reflects all the light and makes our body cool.

• All objects in this world which have a temperature greater than 0 K, emit radiation.
• This radiation energy is proportional to fourth the power of Temperature (T) in Kelvin.
• Q α T⁴
• So, the correct answer is option 4.

• Since 1400 K temperature is much higher temperature compared to the room temperature. So both the polished metal plate and the rough black spot will emit radiation at a higher rate.
• Since the absorptive power of the rough black spot is more compared to the absorptive power of the polished metal plate, so according to Kirchoff's law, the emissive power of the black spot will also be more compared to the polished metal surface.
• So the rough black spot will emit more radiation compared to the polished surface.
• So the spot will appear brighter than the plate.
• Hence, option 1 is correct.