Laws of Radiation MCQ Quiz - Objective Question with Answer for Laws of Radiation - Download Free PDF
Last updated on Apr 17, 2025
Latest Laws of Radiation MCQ Objective Questions
Laws of Radiation Question 1:
What is the approximate value of the Stefan-Boltzmann constant?
Answer (Detailed Solution Below)
Laws of Radiation Question 1 Detailed Solution
Explanation:
Stefan-Boltzmann Constant
- The Stefan-Boltzmann constant, denoted by the symbol σ, is a physical constant that denotes the total energy radiated per unit surface area of a black body in unit time. This radiation is proportional to the fourth power of the black body's thermodynamic temperature.
Stefan-Boltzmann Law: The Stefan-Boltzmann law describes the power radiated from a black body in terms of its temperature. The law is mathematically expressed as:
P = σT4
Where:
- P is the power radiated per unit area.
- σ is the Stefan-Boltzmann constant (5.67 × 10-8 W/m2K4).
- T is the absolute temperature in Kelvin (K).
Importance:
- The Stefan-Boltzmann constant plays a crucial role in thermodynamics and quantum mechanics, particularly in the study of black body radiation.
- It helps in determining the radiant energy emitted by stars, including our Sun, which is essential in astrophysics.
- This constant is used to calculate the radiation heat transfer between objects and is pivotal in various engineering applications like furnace design, radiative cooling of spacecraft, and climate modeling.
Laws of Radiation Question 2:
The Stefan-Boltzmann law, which includes the Stefan-Boltzmann constant, primarily applies to which type of bodies?
Answer (Detailed Solution Below)
Laws of Radiation Question 2 Detailed Solution
Explanation:
Stefan-Boltzmann Law
- The Stefan-Boltzmann law is a fundamental principle in thermal radiation that describes the power radiated from a black body in terms of its temperature. The law is named after physicist Josef Stefan, who experimentally established the law in 1879, and Ludwig Boltzmann, who derived it theoretically.
- The law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature. This implies that even a small increase in temperature results in a significant increase in the amount of radiation emitted.
Mathematical Expression: The Stefan-Boltzmann law is mathematically expressed as:
J = σT4
Where:
- J is the total power radiated per unit surface area of the black body.
- σ is the Stefan-Boltzmann constant, approximately equal to 5.67 × 10-8 W/m2K4.
- T is the absolute temperature of the black body in Kelvin (K).
Option 2: Black bodies
- This option correctly identifies the type of bodies to which the Stefan-Boltzmann law primarily applies. A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Because of this property, a black body is also the best emitter of thermal radiation. The Stefan-Boltzmann law applies specifically to black bodies, which radiate energy according to the formula J = σT4.
Laws of Radiation Question 3:
The maximum energy in the thermal radiation from a hot body A occurs at a wavelength of 130 μm. For a second hot body B, the maximum energy in the thermal radiation occurs at a wavelength of 65μm. The relationship between temperature of A (TA) and temperature of B (TB) is -
Answer (Detailed Solution Below)
Laws of Radiation Question 3 Detailed Solution
Calculation:
We are given that the maximum energy in the thermal radiation from a hot body A occurs at a wavelength of 130 μm. For a second hot body B, the maximum energy in the thermal radiation occurs at a wavelength of 65 μm.
According to Wien's displacement law, the wavelength of the maximum energy emission is inversely proportional to the temperature of the black body, which is given by:
λmax * T = constant
For body A:
λmaxA * TA = constant
For body B:
λmaxB * TB = constant
Since the constant is the same for both bodies, we can write:
λmaxA * TA = λmaxB * TB
Substituting the given wavelengths:
130 μm * TA = 65 μm * TB
Dividing both sides by 65 μm:
2 * TA = TB
Therefore, the relationship between the temperatures is:
Final Answer: TB = 2TA
Laws of Radiation Question 4:
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?
Answer (Detailed Solution Below)
Laws of Radiation Question 4 Detailed Solution
Laws of Radiation Question 5:
The spectral energy density uT(λ) vs wavelength (λ) curve of a black body shows a peak at λ = λmax. If the temperature of the black body is doubled, then
Answer (Detailed Solution Below)
Laws of Radiation Question 5 Detailed Solution
Explanation:
1. Energy Density (u): The formula ( \(u = \sigma T^4 \)) represents the Stefan Boltzmann law, where ( u ) is the energy density, ( \(\sigma\) ) is the Stefan Boltzmann constant, and ( T ) is the temperature. When the temperature is doubled ( \(T \rightarrow 2T \)), the new energy density ( u' ) becomes 16 times the original value, since (\( u' = \sigma (2T)^4 = 16 \sigma T^4\) ).
2. Wien’s Displacement Law (\(λ_m\)): The formula ( \(\lambda_m T = b\) ) represents Wien's displacement law, where ( \(\lambda_m\) ) is the wavelength at the maximum radiation, ( T ) is the temperature, and ( b ) is Wien’s constant. When the temperature doubles ( \(T \rightarrow 2T\) ), the peak wavelength decreases by half ( \(\lambda_m' = \frac{\lambda_m}{2}\) ).
\( \ u = \sigma T^4 \quad \text{as} \quad T \rightarrow 2T \\ u' = \sigma 16T^4 \implies \frac{u'}{u} = 16 \\ \lambda_m T = b \quad \text{as} \quad \lambda_m' 2T = b \\\implies \frac{\lambda_m' 2}{\lambda_m} = 1 \implies \lambda_m' = \frac{\lambda_m}{2} \)
The correct options are (1) and(3) .
Top Laws of Radiation MCQ Objective Questions
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 μm. If the peak wavelength of emitted radiation changes to 2.90 μm, then the temperature (in K) of the black body is
Answer (Detailed Solution Below)
Laws of Radiation Question 6 Detailed Solution
Download Solution PDFConcept:
From Wein's displacement law
\({\lambda _{max}}T = 2898\;\mu m - k\left( {constant} \right)\)
Thus, \({\lambda _{peak}}T = \lambda _{peak}'T'\)
Calculation:
Given,
Black body λpeak = 1.45 μm at 2000 K.
Now, \(\lambda _{peak}' = 2.90\;\mu m\)
1.45 × 2000 = 2.90 × T'
T' = 1000 KAccording to Stefan-Boltzman law the radiation energy emitted byby a black body is directly proportional to (where T is the absolute temperature Of the body)
Answer (Detailed Solution Below)
Laws of Radiation Question 7 Detailed Solution
Download Solution PDFExplanation:
Stephen's Boltzmann law:
According to Stefan’s law, the radiant energy emitted by a black body is directly proportional to the fourth power of its absolute temperature.
Q̇ = єσAT4
Where Q̇ is Radiate energy, σ is the Stefan-Boltzmann Constant, T is the absolute temperature in Kelvin, є is Emissivity of the material, and A is the area of the emitting body.
- Stefan's Law is used to accurately find the temperature Sun, Stars, and the earth.
- A black body is an ideal body that absorbs or emits all types of electromagnetic radiation.
From above it clear that the amount of radiation emitted by a perfectly black body is proportional to the fourth power of temperature on an ideal gas scale. Thus, option 3 is correct.
Thermal radiation extends over the range of
Answer (Detailed Solution Below)
Laws of Radiation Question 8 Detailed Solution
Download Solution PDFConcept:
- Thermal radiation is the transmission of heat in the form of radiant energy from one body to another body by the thermal motion of charged particles of matter.
- The mechanism of radiation is divided into 3 phases
- Conversion of thermal energy of the hot source into electromagnetic waves
- Passage of wave motion through intervening medium
- Transformation of waves into heat
- Thermal radiation is limited to wavelengths ranging from 0.1 to 100 µm of the electromagnetic spectrum.
- It includes some portion of UV radiation (0.1 to 0.4 µm), entire visible radiation (0.4 to 0.7 µm), and entire infrared radiation (0.7 to 100 µm)
According to Stefan-Boltzmann Law, the emissive power is directly proportional to the fourth power of its absolute temperature. This applies to
Answer (Detailed Solution Below)
Laws of Radiation Question 9 Detailed Solution
Download Solution PDFExplanation:
Stefen's Boltzmann law:
According to Stefan’s law, the radiant energy emitted by a black body is directly proportional to the fourth power of its absolute temperature.
P = σAT4
Where P is Radiate energy, σ is the Stefan-Boltzmann Constant, T is the absolute temperature in Kelvin, є is Emissivity of the material, and A is Area of the emitting body.
- Stefan's Law is used to accurately find the temperature Sun, Stars, and the earth.
- A black body is an ideal body that absorbs or emits all types of electromagnetic radiation.
Additional Information
If anybody radiates equally in all directions, then it is called a diffuse body.
For diathermonous bodies also known as transparent bodies, the transmissivity (τ) = 1
⇒ absorptivity (α) = 0; reflectivity (ρ) = 0;
For perfect black body,
- Absorptivity = 1; Emissivity = 1;
- Transmissivity = 0; Reflectivity = 0;
Gray body
- If a body absorbs or emits radiation in constant proportions irrespective of wavelength then it is called a Gray body.
- Absorptivity = Transmissivity = Reflectivity = constant
Opaque body
- For opaque bodies, Transmissivity = 0 ⇒ absorptivity + reflectivity = 1;
White body
- For white bodies, reflectivity = 1 ⇒ transmissivity = 0; absorptivity = 0;
A ________ body reflects entire radiation incident on it.
Answer (Detailed Solution Below)
Laws of Radiation Question 10 Detailed Solution
Download Solution PDFExplanation:
White Body:
- A body that is called a white body is one that reflects almost all radiations incident upon it and does not absorb or transmit any part of it.
- For a white body, α = τ = 0, ρ = 1
Additional Information
Black body
- 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 all the incident radiation 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)
- Transparent body: τ = 1, α = ρ = 0
- Opaque body: τ = 0
According to Planck’s law, the wavelength corresponding to the maximum energy is proportional to
Answer (Detailed Solution Below)
Laws of Radiation Question 11 Detailed Solution
Download Solution PDFExplanation:
According to Planck's law, the wavelength corresponding to the maximum energy is inversely proportional to Temperature. It is obtained from Wien's displacement law. The Wien’s displacement law can be obtained by determining the maxima of Planck’s law.
Wein’s displacement law:
- According to this law, the radiation curve for a black body is different at different temperatures and peak wavelengths are inversely proportional to the temperature of the body.
- Maximum wavelength = \(\lambda ~=~\frac{b}{T}\)
Where b is Wein's constant = 3 × 10-3 m.K and T is the temperature of the body (In kelvin unit).
The temperature of a surface with 0.2 m2 area is 17 deg C. Calculate the wavelength corresponding to maximum monochromatic emissive power
Answer (Detailed Solution Below)
Laws of Radiation Question 12 Detailed Solution
Download Solution PDFConcept:
Wein’s displacement law gives the relationship between wavelength and temperature.
λpeak ⋅ T = 2.898 × 10-3 m ⋅ K
λpeak ⋅ T = 2898 micrometre ⋅ K
Calculation:
Given,T = 17°C = 290 K
λpeak × 290 = 2898
Which of the following is not the characteristics of Planck’s black body radiation distribution
Answer (Detailed Solution Below)
Laws of Radiation Question 13 Detailed Solution
Download Solution PDFExplanation:
Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T.
Planck’s law for the energy Eλ radiated per unit volume by a cavity of a blackbody in the wavelength interval λ to λ + Δλ can be written in terms of Planck’s constant (h), the speed of light (c = λ × v), the Boltzmann constant (k), and the absolute temperature (T):
Energy per unit volume per unit wavelength:
\({E_\lambda } = \frac{{8\pi hc}}{{{\lambda ^5}}} \times \frac{1}{{{e^{\frac{{hc}}{{kT\lambda }} - 1}}}}\)
Energy per unit volume per unit frequency:
\({E_\nu } = \frac{{8\pi h}}{{{c^3}}} \times \frac{{{\nu ^3}}}{{{e^{\frac{{hv}}{{kT}} - 1}}}}\)
So Planck’s distribution function:
\(E\left( {\omega ,T} \right) = \frac{1}{{{e^{\frac{{h\omega }}{\tau }}} - 1}}\)
Using planck’s law, when we plot Ebλ with λ, we get the curve as shown below.
As temperature increases, the peak of the curve shift towards a lower wavelength.
The heat transfer equation Q = σAT4 is called
Answer (Detailed Solution Below)
Laws of Radiation Question 14 Detailed Solution
Download Solution PDFExplanation:
Stephen's Boltzmann law:
According to Stefan’s law, the radiant energy emitted by a black body is directly proportional to the fourth power of its absolute temperature.
P = σAT4
Where P is Radiate energy, σ is the Stefan-Boltzmann Constant, T is the absolute temperature in Kelvin, є is Emissivity of the material, and A is Area of the emitting body.
- Stefan's Law is used to accurately find the temperature Sun, Stars, and the earth.
- A black body is an ideal body that absorbs or emits all types of electromagnetic radiation.
Additional Information
Newton's law of cooling
According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body, and its surroundings.
Rate of cooling ∝ ΔT
\(\frac{{dT}}{{dt}} = - k\left( {T - {T_\infty }} \right)\)
Fourier law
According to Fourier law "rate of heat flow by conduction is proportional to the area measured normal to the direction of heat flow, and to the temperature gradient in that direction".
\(Q = -kA\frac{{dT}}{{dx}}\)
where Q = rate of heat flow, k = proportionality constant which is the thermal conductivity of the material, A = cross-sectional area perpendicular to the direction of heat flow, \(\frac{dT}{dx}\) = temperature gradient in the direction of heat flow
Here negative sign shows that heat always flows from higher temperature to lower temperature, means the temperature gradient will always be negative in the direction of heat flow.
The spectral energy distribution for a body at temperature T is shown in figure. Now it the temperature of the body is increased
Answer (Detailed Solution Below)
Laws of Radiation Question 15 Detailed Solution
Download Solution PDFCONCEPT:
Wien's Displacement Law:
- According to Wien's law the product of wavelength corresponding to the maximum intensity of radiation and temperature of the body (in Kelvin) is constant, i.e.
λmT = b = constant
Where b = Wien's constant and has value 2.89 10-3 m-K.
EXPLANATION:
- As the temperature of the body increases, the wavelength at which the spectral intensity (Eλ) is maximum shifts towards left. Therefore option 2 is correct.
- Therefore it is also called Wien's displacement law.