Black body radiation MCQ Quiz - Objective Question with Answer for Black body radiation - Download Free PDF

Let \(A_2\) be the area when length and breadth are reduced by half. Thus the area will be \(\dfrac {1}{4}\)th of \(A_1\)

\(\therefore A_2=\dfrac {1}{4} \times 32=8\)

Given \(T_1=127^\circ \text{C}=400 \, \text{K}\)

Given \(T_2=327^\circ \text{C}=600 \, \text{K}\)

From Stefan's law \(E=\sigma AT^4\)

\(\therefore \dfrac{E_1}{E_2}= \dfrac {A_1{T_1}^4}{A_2{T_2}^4}=\dfrac {32 \times (400)^4}{8 \times (600)^4}=\dfrac{64}{81}\)

\(\therefore \dfrac{E_2}{E_1}= \dfrac{81}{64}\)

Concept Used:

The emissivity (e) of a surface is related to its reflectivity (r) and absorptivity (a) based on Kirchhoff's Law of Radiation:

e = a

Since energy conservation applies, we use the relation:

r + a = 1

where:

r = Reflectivity of the surface

a = Absorptivity of the surface

e = Emissivity of the surface

Calculation:

Given that the reflectivity of the surface is r = 0.7 , we calculate emissivity:

⇒ a = 1 - r

⇒ a = 1 - 0.7

⇒ a = 0.3

From Kirchhoff's Law:

⇒ e = a = 0.3

Looking at Column-3 , the corresponding value for 0.3 is (R).

From Column-1 and Column-2 , Kirchhoff's Law states that:

⇒ Absorptivity equals emissivity in thermal equilibrium → (III) (iii)

Thus, the correct match is (III) (iii) (R) .

Correct Option: Option 4

Let \(A_2\) be the area when length and breadth are reduced by half. Thus the area will be \(\dfrac {1}{4}\)th of \(A_1\)

\(\therefore A_2=\dfrac {1}{4} \times 32=8\)

Given \(T_1=127^\circ \text{C}=400 \, \text{K}\)

Given \(T_2=327^\circ \text{C}=600 \, \text{K}\)

From Stefan's law \(E=\sigma AT^4\)

\(\therefore \dfrac{E_1}{E_2}= \dfrac {A_1{T_1}^4}{A_2{T_2}^4}=\dfrac {32 \times (400)^4}{8 \times (600)^4}=\dfrac{64}{81}\)

\(\therefore \dfrac{E_2}{E_1}= \dfrac{81}{64}\)

Answer : 2

Solution :

From Wien's Displacement law,

λ_max = \(\frac{b}{T} \Rightarrow T=\frac{b}{\lambda_{\max }}\)

From Stefan-Boltzmann law

E = σT¹ = σ\(\left(\frac{b}{\lambda_{\max }}\right)^{4}\)

Let the new emissive power be E'.

∴ E' = \(\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}\right)^{4}\)

E' = \(\frac{81}{16} E\)

Concept:

Wien's Displacement Law and Stefan-Boltzmann Law:

• Wien's displacement law states that the wavelength at which a black body radiates maximum energy is inversely proportional to its temperature:
• λₘT = b, where λₘ is the wavelength of maximum emission, T is the temperature, and b is Wien's constant.
• The Stefan-Boltzmann law gives the total emissive power of a black body as E = σT⁴, where σ is the Stefan-Boltzmann constant.
• When the wavelength of maximum energy changes from λ to 2λ/3, the temperature will change accordingly due to the relation in Wien’s law.

Calculation:

Let the initial temperature be T₁, and initial wavelength of maximum energy be λ.

From Wien’s displacement law, λₘT₁ = b, so T₁ = b / λ.

For the new temperature, let it be T₂, and the new wavelength of maximum energy is 2λ/3. Thus, from Wien’s law, λₘT₂ = b, so T₂ = b / (2λ/3) = 3b / 2λ.

The ratio of temperatures is T₂ / T₁ = (3b / 2λ) / (b / λ) = 3/2.

The emissive power is proportional to the fourth power of temperature: E₂ / E₁ = (T₂ / T₁)⁴ = (3/2)⁴ = 81/16.

∴ The emissive power is 81/16.
Hence, the correct option is 1).

• The good absorbers of heat are good-emitters of heat.

Key Points

• Concept
  • The fact that 'good absorbers of heat are good emitters' is based on Kirchoff's radiation law.
  • The Kirchoff's radiation law states that at a given temperature the coefficient of absorption of a body is equal to its coefficient of emission.
• Explanation
  • 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.

Additional Information

• Application of Kirchoff's Radiation Law,
  • Sand is rough black, so it is a good absorber and hence, in deserts, days will be very hot.
  • A person with black skin experiences more heat and cold as compared to white skin person.
  • When a green glass is heated in a furnace and taken out, it is found to glow with red light, because red and green are complementary colours.

Concept:

The amount of electrical energy transferred to an appliance depends on its power and the duration of time it is used.

Electrical energy is measured in kilowatt-hours, kWh. One unit is 1 kWh.

Electrical energy transferred is given as,

E = P × t

Where, P is the power in kilowatts (kW),  T is the time in hours (h)

Calculation:

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.

Concept:

Stefan-Boltzmann Law:

The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:

E ∝ T4

E = σT4

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

Given:

T₁ = 27°C ⇒ 300 K and

T1 = 927°C ⇒  1200 K

E = σT4

\(\frac{E_1}{E_2}=\frac{T_1^4}{T_2^4}\)

\(\frac{E_1}{E_2}=\frac{300^4}{1200^4}=\frac{1}{4^4}=\frac{1}{256}\)

option(2)

CONCEPT:

• Black body: Those type of body that neither reflect nor transmits but absorbs the whole of the heat radiation called the black body.
  • Energy distribution of black body radiation state that when wavelength increases, the energy emitted from the black body radiation will also be increasing.

EXPLANATION:

• 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.

Concept:

The waves which have both electric as well as magnetic field such that electric and magnetic fields are perpendicular to each other and also to direction of propagation is called as electromagnetic waves.

Calculation:

Intensity of light =20 W/cm2 

Area = 20 cm2

Time = 1 minute = 60 seconds

Energy received = Intensity × Area × Time

= 20 × 20 × 60

= 24 × 10³ J

CONCEPT:

• A black body is an object that absorbs all the radiant energy reaching its surface.
• No actual body is perfectly black; the concept of a black body is an idealization with which the radiation characteristics of real bodies can be compared.

Properties of the black body:

• It absorbs the incident radiation of all wavelength falling on it and does not transmit or reflect regardless of wavelength and direction
• It emits the maximum amount of thermal radiation at all wavelengths at any specified temperature
• It is a diffuser emitter (i.e. the radiation emitted by a black body is independent of direction)

EXPLANATION:

• 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.

Concept: 

• Every object absorbs some amount of radiation that falls on it.
• A blackbody absorbs all the radiation falling on it. 
• At a certain temperature, it emits radiation absorbed.
• It emits radiation of all the wavelengths.

Explanation: 

• Wein's displacement law: Emitted wavelength peak is inversely proportional to the temperature of a blackbody.
• As the temperature of a blackbody increases the highest peak of radiated wavelength moves towards a shorter wavelength.

λmax α 1 / T

λ_max = b / T

b is Wein's displacement constant

• In given condition λ_max1 = 1.56 μm at T₁ = 2000K

λ_max2 = 1.8 μm T₂ = ?

\(\frac {\lambda_{max1}}{\lambda_{max2}} = \frac{T_2}{T_1}\)

\(\frac{1.56 * 10^{-6}}{1.8*10^{-6}}=\frac{T_2}{2000}\)

\(T_2= \frac{2000*1.56}{1.8}=\frac{3120}{1.8}\)

\(T_2 = 1733 \)  K

Therefore option 3 is correct

Additional Information

•  Wein's displacement law is useful to calculate the temperature of stellar objects like stars.  
• When metal is heated at a higher temperature it initially turns red,
• As temperature increases, it appears orange and then yellow at maximum temperature it appears white.
• This shows as temperature increases wavelength becomes shorter.

CONCEPT:

• In the world, all objects emit radiation if they have a temperature greater than absolute zero ( 0K ).
• And this radiation energy (Q) is proportional to fourth the power of Temperature (T) in Kelvin.

i.e. Q α T4 or mathematically  Q = σAT4

where σ is Stefan-Boltzmann constant and A is the area from which radiation is emitted. 

EXPLANATION:

• 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.

• An object that absorbs and emits all possible radiation with 100% is called a blackbody.

CONCEPT

Black body Radiation: A body that absorbs all the radiations falling on it is called a blackbody. This body will emit radiation at its fastest rate is called blackbody radiation. A blackbody is also called an ideal radiator.

EXPLANATION

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.

option(2)

CONCEPT:

• White and shiny surfaces are the poor absorber of heat and light where dark and black surfaces are the good absorber of heat and light.
• A person with dark skin will experience more heat and mole cold in comparison to white skin people.

EXPLANATION:

• 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.

Additional Information

• Dark people like people who live in Africa have the ability to absorb, an average of 36% heat or solar radiation than any other people in the world. 
• There are no such bodies that are perfectly absorbing or emitting light, this means no light can be absorbed or emit totally.