To calculate the ratio of the surface temperatures of the Sun and star X, we use Wien’s displacement law: λ_max × T = b where: - λ_max is the wavelength at maximum radiation, - T is the temperature of the object, - b is Wien's constant (2.9 × 10⁶ nm·K). The ratio of the temperatures is given by: (T_Read more
To calculate the ratio of the surface temperatures of the Sun and star X, we use Wien’s displacement law:
λ_max × T = b
where:
– λ_max is the wavelength at maximum radiation,
– T is the temperature of the object,
– b is Wien’s constant (2.9 × 10⁶ nm·K).
The ratio of the temperatures is given by:
(T_sun / T_star_X) = λ_max_star_X / λ_max_sun
Substitute the given values:
– λ_max_sun = 510 nm,
– λ_max_star_X = 350 nm.
(T_sun / T_star_X) = 350 / 510 = 0.686
Thus, the ratio of surface temperatures is approximately 0.68.
We can apply the Stefan-Boltzmann law to compute the change in the total energy emitted: E = σ T⁴ A Here, - E is the total energy emitted - σ is the Stefan-Boltzmann constant - T is the surface temperature - A is the surface area of the object Surface area of sphere is given as: A = 4π r² If the radRead more
We can apply the Stefan-Boltzmann law to compute the change in the total energy emitted:
E = σ T⁴ A
Here,
– E is the total energy emitted
– σ is the Stefan-Boltzmann constant
– T is the surface temperature
– A is the surface area of the object
Surface area of sphere is given as:
A = 4π r²
If the radius of the sun increases by a factor of 100 and the temperature decreases by half, we need to calculate the change in energy.
Let the initial energy be E₁, and the final energy be E₂.
Initial energy: E₁ = σ T₁⁴ (4π r₁²)
Final energy: E₂ = σ (T₁/2)⁴ (4π (100r₁)²)
Simplifying the ratio of the final energy to the initial energy:
(E₂ / E₁) = (T₁/2)⁴ (100²) = (1/16) (10000) = 625
Thus, the total energy emitted increases by a factor of 625.
According to the Stefan-Boltzmann law, the energy radiated by a black body is given by: E = σ T⁴ where: - E is the radiated energy, - σ is the Stefan-Boltzmann constant, - T is the temperature in Kelvin. Now, when the temperature is reduced to T/2, the new energy radiated will be: E' = σ (T/2)⁴ = (σRead more
According to the Stefan-Boltzmann law, the energy radiated by a black body is given by:
E = σ T⁴
where:
– E is the radiated energy,
– σ is the Stefan-Boltzmann constant,
– T is the temperature in Kelvin.
Now, when the temperature is reduced to T/2, the new energy radiated will be:
E’ = σ (T/2)⁴ = (σ T⁴) / 16
So the radiated energy decreases by a factor of 16, meaning the new radiated energy is E/16.
According to the Stefan-Boltzmann law, the radiated energy of a black body is proportional to the fourth power of its temperature: E ∝ T⁴ Let E₁ be the energy radiated at temperature T₁ = 150°C = 150 + 273 = 423 K, and let E₂ be the energy radiated at T₂ = 300°C = 300 + 273 = 573 K. We know that: E₁Read more
According to the Stefan-Boltzmann law, the radiated energy of a black body is proportional to the fourth power of its temperature:
E ∝ T⁴
Let E₁ be the energy radiated at temperature T₁ = 150°C = 150 + 273 = 423 K, and let E₂ be the energy radiated at T₂ = 300°C = 300 + 273 = 573 K.
Black holes orbiting normal stars are observed in the X-ray region of the electromagnetic spectrum because of a process called accretion. Here is how it works in detail: 1. Matter Transfer from the Star: In a binary system in which a black hole orbits a normal star, the intense gravitational force eRead more
Black holes orbiting normal stars are observed in the X-ray region of the electromagnetic spectrum because of a process called accretion. Here is how it works in detail:
1. Matter Transfer from the Star:
In a binary system in which a black hole orbits a normal star, the intense gravitational force exerted by the black hole could strip away material from its companion star. This is very common in systems where the star expands or the two stars are close enough for a large outer layer to be pulled into the black hole.
2. Accretion Disk Formation:
Since matter drawn from the star does not collapse directly inside a black hole, but instead begins spiraling because of angular momentum, it can eventually form an incredibly dense, hot **accretion disk.**
3. Heating of the Accretion Disk:
As the gas in the accretion disk spirals closer to the black hole, it gets compressed and heated up to millions of degrees Kelvin. At such incredibly high temperatures, the gas radiates electromagnetic waves.
4. Emission in the X-ray Region:
This radiation peaks in the X-ray region of the spectrum, since temperatures are high enough to excite particles to X-ray energies. X-rays can escape the black hole’s neighborhood and be observed by telescopes built to see high-energy radiation.
5. Why Not Other Regions?
– Visible Region: The radiation of the accretion disk in the visible range is rather weak and can be overpowered by the light of the star itself.
– Microwave Region: The disk emits little radiation in the low-energy microwave range.
– Gamma-Ray Region: Gamma rays are produced in less frequent, more energetic events like jets or near collisions with black holes but do not make up the dominant region in normal binary systems.
Conclusion:
The presence of X-rays in binary systems provides strong evidence of a black hole because normal stars and most stellar phenomena do not emit such intense X-rays. Observatories like Chandra X-ray Observatory and XMM-Newton specialize in detecting these X-rays to study black holes.
Prevost's Theory of Heat Exchange : Prevost's theory of heat exchange is that all bodies emit continuous thermal radiation; it doesn't matter what the temperature is or what their surroundings are. Emission depends on a body's nature and temperature, not whether or not there are other bodies around.Read more
Prevost’s Theory of Heat Exchange :
Prevost’s theory of heat exchange is that all bodies emit continuous thermal radiation; it doesn’t matter what the temperature is or what their surroundings are. Emission depends on a body’s nature and temperature, not whether or not there are other bodies around. In contrast to the emission, bodies absorb radiation from their surroundings.
Heat exchange happens between two bodies because of the difference in temperatures of those bodies. The hotter body radiates more energy than it absorbs, whereas the cooler body absorbs more energy than it radiates. This means that there is a net transfer of heat from the hotter body towards the cooler. When a body is in thermal equilibrium with the surroundings, it radiates as well as absorbs energy at the same rate. In this case, there is no net transfer of heat. This process is dynamic, as radiation as well as absorption is in continuous motion. Good Absorbers are Good Radiators :
By Prevost’s theory, a good absorber is also a good radiator.
This is because the properties that enable a surface to absorb energy efficiently allow it also to emit energy efficiently. For instance, black surfaces, which are excellent absorbers of radiation, are also excellent radiators. The relationship is quantified by Kirchhoff’s Law, which asserts that the emissive power of a body is proportional to its absorptive power at the same temperature and wavelength.
Experimental Observations Relevant to Kirchhoff's Law of Heat Radiation: Kirchhoff's law of heat radiation states that for a body in thermal equilibrium, the ratio of its emissive power to its absorptive power is constant and equal to the emissive power of a perfect blackbody at the same temperatureRead more
Experimental Observations Relevant to Kirchhoff’s Law of Heat Radiation:
Kirchhoff’s law of heat radiation states that for a body in thermal equilibrium, the ratio of its emissive power to its absorptive power is constant and equal to the emissive power of a perfect blackbody at the same temperature. Several experimental observations confirm this law:
1. Emission and absorption by black bodies:
Black bodies are perfect absorbers and thus also the best emitters of radiation. Experiments on cavity radiators, that approximate black bodies, demonstrate that their radiation spectrum does indeed behave according to theory, for example Planck’s law, as demanded by Kirchhoff’s principle.
2. Properties of surfaces and radiation:
Blackened surfaces absorb more radiation and emit more heat, while polished surfaces, which are poor absorbers, are also poor emitters. Comparative experiments confirm the law’s validity.
3. Absorption lines in stellar spectra:
Kirchhoff’s law accounts for the absorption lines that appear in stellar spectra. These lines arise from gases in the atmosphere of the star absorbing certain wavelengths of light emitted by the hotter interior, which proves that a good absorber is also a good emitter.
4. Cooling and heating experiments:
Objects with high absorptivity, such as black surfaces, cool faster because they radiate heat more efficiently than reflective surfaces.
5. Greenhouse effect research:
The gases like carbon dioxide and water vapor are able to absorb infrared radiation strongly as well as radiate heat effectively. These observations have been established experimentally, which are central to the studies on climate.
The above results clearly illustrate that a body which absorbs and emits radiations have reciprocal characteristics.
Latent heat of vaporization is about the change in phase from liquid to gas; it involves breaking almost all the intermolecular bonds. Such a process needs much more energy. - The latent heat of fusion involves the change from the solid state to liquid, where just some intermolecular bonds are releaRead more
Latent heat of vaporization is about the change in phase from liquid to gas; it involves breaking almost all the intermolecular bonds. Such a process needs much more energy.
– The latent heat of fusion involves the change from the solid state to liquid, where just some intermolecular bonds are released; therefore, less energy is required.
– Usually, the latent heat of vaporization is much higher than the latent heat of fusion for the same substance.
Explanation: As the temperature of a substance is increased, it emits thermal radiation, and its color changes because the radiation spectrum shifts towards shorter wavelengths: 1. When the temperature is lower, the substance emits infrared radiation, which is not visible. 2. With increasing temperaRead more
Explanation:
As the temperature of a substance is increased, it emits thermal radiation, and its color changes because the radiation spectrum shifts towards shorter wavelengths:
1. When the temperature is lower, the substance emits infrared radiation, which is not visible.
2. With increasing temperature, it begins to emit red light which is the first coloured light.
3. The glow now becomes yellow with further heating and eventually goes white when the emitted light ranges over a bigger spectrum, covering the whole visible wavelengths.
This happens to be due to blackbody radiation and Planck’s law in which higher temperatures correspond to higher-energy radiation while, at the same time, there is a shift in the emission spectrum.
Wien's Displacement Law states that the wavelength (λ_max) at which the intensity of radiation emitted by a black body is maximum is inversely proportional to its absolute temperature (T). In other words, as the temperature of the black body increases, the wavelength at which it emits the most radiaRead more
Wien’s Displacement Law states that the wavelength (λ_max) at which the intensity of radiation emitted by a black body is maximum is inversely proportional to its absolute temperature (T). In other words, as the temperature of the black body increases, the wavelength at which it emits the most radiation decreases.
Mathematically, Wien’s Displacement Law is given as:
λ_max = b / T
where:
– λ_max is the wavelength at which the emission intensity is maximum,
– T is the absolute temperature of the black body in Kelvin (K),
– b is Wien’s displacement constant, approximately 2.898 × 10⁻³ m·K.
Illustration:
– Object at 300 K: Using Wien’s Displacement Law, we calculate the wavelength where its radiation peaks.
λ_max = (2.898 × 10⁻³) / 300 = 9.66 μm
This is in the infrared region.
– Object at 600 K: The peak wavelength of emission is twice the object at 300 K since the temperature has an inverse proportionality to peak wavelength.
λ_max = (2.898 × 10⁻³) / 600 = 4.83 μm
It is still within the infrared, but shorter now.
Relevance of Wien’s Displacement Law:
1. Temperature Determination from Radiation:
Temperature of any object could be determined by the peak wavelength of emitted radiation under the law. It finds many applications in astrophysics, for example, to estimate the temperature of stars as a function of the color of stars. Therefore, it gives a color equivalent estimation.
2. Color of Stars
According to Wien’s law, hotter stars emit more radiation at shorter wavelengths, causing them to appear bluer, and cooler stars emit at longer wavelengths, making them appear redder.
3. Thermal Radiation Understanding:
Wien’s law helps us understand the nature of thermal radiation and how temperature affects the radiation emitted by objects, which is vital in many fields, including climatology and energy studies.
4. Temperature Measurement Applications: The law is applied in infrared thermometry, which allows objects to be measured for temperature without direct contact. This is found in industrial, medical, and scientific applications.
The sun emits a light with maximum wavelength 510 nm, while another star X emits a light with maximum wavelength of 350nm. What is the ratio surface temperature of the sun and the star X?
To calculate the ratio of the surface temperatures of the Sun and star X, we use Wien’s displacement law: λ_max × T = b where: - λ_max is the wavelength at maximum radiation, - T is the temperature of the object, - b is Wien's constant (2.9 × 10⁶ nm·K). The ratio of the temperatures is given by: (T_Read more
To calculate the ratio of the surface temperatures of the Sun and star X, we use Wien’s displacement law:
λ_max × T = b
where:
– λ_max is the wavelength at maximum radiation,
– T is the temperature of the object,
– b is Wien’s constant (2.9 × 10⁶ nm·K).
The ratio of the temperatures is given by:
(T_sun / T_star_X) = λ_max_star_X / λ_max_sun
Substitute the given values:
– λ_max_sun = 510 nm,
– λ_max_star_X = 350 nm.
(T_sun / T_star_X) = 350 / 510 = 0.686
Thus, the ratio of surface temperatures is approximately 0.68.
The correct answer is 0.68.
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Suppose the sun expands so that its radius becomes 100 times its present radius and its surface temperature becomes half of its present value. The total energy emitted by it then will increase by a factor of
We can apply the Stefan-Boltzmann law to compute the change in the total energy emitted: E = σ T⁴ A Here, - E is the total energy emitted - σ is the Stefan-Boltzmann constant - T is the surface temperature - A is the surface area of the object Surface area of sphere is given as: A = 4π r² If the radRead more
We can apply the Stefan-Boltzmann law to compute the change in the total energy emitted:
E = σ T⁴ A
Here,
– E is the total energy emitted
– σ is the Stefan-Boltzmann constant
– T is the surface temperature
– A is the surface area of the object
Surface area of sphere is given as:
A = 4π r²
If the radius of the sun increases by a factor of 100 and the temperature decreases by half, we need to calculate the change in energy.
Let the initial energy be E₁, and the final energy be E₂.
Initial energy: E₁ = σ T₁⁴ (4π r₁²)
Final energy: E₂ = σ (T₁/2)⁴ (4π (100r₁)²)
Simplifying the ratio of the final energy to the initial energy:
(E₂ / E₁) = (T₁/2)⁴ (100²) = (1/16) (10000) = 625
Thus, the total energy emitted increases by a factor of 625.
The correct answer is 625.
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A black body at a high temperature T K radiates energy at the rate of E Wm⁻². When the temperature falls to T/2 K, the radiated energy will be
According to the Stefan-Boltzmann law, the energy radiated by a black body is given by: E = σ T⁴ where: - E is the radiated energy, - σ is the Stefan-Boltzmann constant, - T is the temperature in Kelvin. Now, when the temperature is reduced to T/2, the new energy radiated will be: E' = σ (T/2)⁴ = (σRead more
According to the Stefan-Boltzmann law, the energy radiated by a black body is given by:
E = σ T⁴
where:
– E is the radiated energy,
– σ is the Stefan-Boltzmann constant,
– T is the temperature in Kelvin.
Now, when the temperature is reduced to T/2, the new energy radiated will be:
E’ = σ (T/2)⁴ = (σ T⁴) / 16
So the radiated energy decreases by a factor of 16, meaning the new radiated energy is E/16.
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A metal rod at a temperature of 150°C, radiates energy at a rate of 20 W. If its temperature is increased to 300°C, then it will radiate at the rate of
According to the Stefan-Boltzmann law, the radiated energy of a black body is proportional to the fourth power of its temperature: E ∝ T⁴ Let E₁ be the energy radiated at temperature T₁ = 150°C = 150 + 273 = 423 K, and let E₂ be the energy radiated at T₂ = 300°C = 300 + 273 = 573 K. We know that: E₁Read more
According to the Stefan-Boltzmann law, the radiated energy of a black body is proportional to the fourth power of its temperature:
E ∝ T⁴
Let E₁ be the energy radiated at temperature T₁ = 150°C = 150 + 273 = 423 K, and let E₂ be the energy radiated at T₂ = 300°C = 300 + 273 = 573 K.
We know that:
E₁ / E₂ = (T₁ / T₂)⁴
Substitute the values:
20 / E₂ = (423 / 573)⁴
Solving for E₂:
E₂ = 20 × (573 / 423)⁴ ≈ 37.2 W
The correct answer is 37.2 W.
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Black holes in orbit around a normal star are detected from
Black holes orbiting normal stars are observed in the X-ray region of the electromagnetic spectrum because of a process called accretion. Here is how it works in detail: 1. Matter Transfer from the Star: In a binary system in which a black hole orbits a normal star, the intense gravitational force eRead more
Black holes orbiting normal stars are observed in the X-ray region of the electromagnetic spectrum because of a process called accretion. Here is how it works in detail:
1. Matter Transfer from the Star:
In a binary system in which a black hole orbits a normal star, the intense gravitational force exerted by the black hole could strip away material from its companion star. This is very common in systems where the star expands or the two stars are close enough for a large outer layer to be pulled into the black hole.
2. Accretion Disk Formation:
Since matter drawn from the star does not collapse directly inside a black hole, but instead begins spiraling because of angular momentum, it can eventually form an incredibly dense, hot **accretion disk.**
3. Heating of the Accretion Disk:
As the gas in the accretion disk spirals closer to the black hole, it gets compressed and heated up to millions of degrees Kelvin. At such incredibly high temperatures, the gas radiates electromagnetic waves.
4. Emission in the X-ray Region:
This radiation peaks in the X-ray region of the spectrum, since temperatures are high enough to excite particles to X-ray energies. X-rays can escape the black hole’s neighborhood and be observed by telescopes built to see high-energy radiation.
5. Why Not Other Regions?
– Visible Region: The radiation of the accretion disk in the visible range is rather weak and can be overpowered by the light of the star itself.
– Microwave Region: The disk emits little radiation in the low-energy microwave range.
– Gamma-Ray Region: Gamma rays are produced in less frequent, more energetic events like jets or near collisions with black holes but do not make up the dominant region in normal binary systems.
Conclusion:
The presence of X-rays in binary systems provides strong evidence of a black hole because normal stars and most stellar phenomena do not emit such intense X-rays. Observatories like Chandra X-ray Observatory and XMM-Newton specialize in detecting these X-rays to study black holes.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
Write the main features of Prevost’s theory of heat exchange. How does this theory lead to the fact that good absorbers are good radiators?
Prevost's Theory of Heat Exchange : Prevost's theory of heat exchange is that all bodies emit continuous thermal radiation; it doesn't matter what the temperature is or what their surroundings are. Emission depends on a body's nature and temperature, not whether or not there are other bodies around.Read more
Prevost’s Theory of Heat Exchange :
Prevost’s theory of heat exchange is that all bodies emit continuous thermal radiation; it doesn’t matter what the temperature is or what their surroundings are. Emission depends on a body’s nature and temperature, not whether or not there are other bodies around. In contrast to the emission, bodies absorb radiation from their surroundings.
Heat exchange happens between two bodies because of the difference in temperatures of those bodies. The hotter body radiates more energy than it absorbs, whereas the cooler body absorbs more energy than it radiates. This means that there is a net transfer of heat from the hotter body towards the cooler. When a body is in thermal equilibrium with the surroundings, it radiates as well as absorbs energy at the same rate. In this case, there is no net transfer of heat. This process is dynamic, as radiation as well as absorption is in continuous motion. Good Absorbers are Good Radiators :
By Prevost’s theory, a good absorber is also a good radiator.
This is because the properties that enable a surface to absorb energy efficiently allow it also to emit energy efficiently. For instance, black surfaces, which are excellent absorbers of radiation, are also excellent radiators. The relationship is quantified by Kirchhoff’s Law, which asserts that the emissive power of a body is proportional to its absorptive power at the same temperature and wavelength.
Click here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
Describe some experimental observations to which Kirchhoff’s law of heat radiation is applicable.
Experimental Observations Relevant to Kirchhoff's Law of Heat Radiation: Kirchhoff's law of heat radiation states that for a body in thermal equilibrium, the ratio of its emissive power to its absorptive power is constant and equal to the emissive power of a perfect blackbody at the same temperatureRead more
Experimental Observations Relevant to Kirchhoff’s Law of Heat Radiation:
Kirchhoff’s law of heat radiation states that for a body in thermal equilibrium, the ratio of its emissive power to its absorptive power is constant and equal to the emissive power of a perfect blackbody at the same temperature. Several experimental observations confirm this law:
1. Emission and absorption by black bodies:
Black bodies are perfect absorbers and thus also the best emitters of radiation. Experiments on cavity radiators, that approximate black bodies, demonstrate that their radiation spectrum does indeed behave according to theory, for example Planck’s law, as demanded by Kirchhoff’s principle.
2. Properties of surfaces and radiation:
Blackened surfaces absorb more radiation and emit more heat, while polished surfaces, which are poor absorbers, are also poor emitters. Comparative experiments confirm the law’s validity.
3. Absorption lines in stellar spectra:
Kirchhoff’s law accounts for the absorption lines that appear in stellar spectra. These lines arise from gases in the atmosphere of the star absorbing certain wavelengths of light emitted by the hotter interior, which proves that a good absorber is also a good emitter.
4. Cooling and heating experiments:
Objects with high absorptivity, such as black surfaces, cool faster because they radiate heat more efficiently than reflective surfaces.
5. Greenhouse effect research:
The gases like carbon dioxide and water vapor are able to absorb infrared radiation strongly as well as radiate heat effectively. These observations have been established experimentally, which are central to the studies on climate.
The above results clearly illustrate that a body which absorbs and emits radiations have reciprocal characteristics.
Click here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
The latent heat of vaporisation of a substance is always
Latent heat of vaporization is about the change in phase from liquid to gas; it involves breaking almost all the intermolecular bonds. Such a process needs much more energy. - The latent heat of fusion involves the change from the solid state to liquid, where just some intermolecular bonds are releaRead more
Latent heat of vaporization is about the change in phase from liquid to gas; it involves breaking almost all the intermolecular bonds. Such a process needs much more energy.
– The latent heat of fusion involves the change from the solid state to liquid, where just some intermolecular bonds are released; therefore, less energy is required.
– Usually, the latent heat of vaporization is much higher than the latent heat of fusion for the same substance.
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On increasing the temperature of a substance gradually, its colour becomes
Explanation: As the temperature of a substance is increased, it emits thermal radiation, and its color changes because the radiation spectrum shifts towards shorter wavelengths: 1. When the temperature is lower, the substance emits infrared radiation, which is not visible. 2. With increasing temperaRead more
Explanation:
As the temperature of a substance is increased, it emits thermal radiation, and its color changes because the radiation spectrum shifts towards shorter wavelengths:
1. When the temperature is lower, the substance emits infrared radiation, which is not visible.
2. With increasing temperature, it begins to emit red light which is the first coloured light.
3. The glow now becomes yellow with further heating and eventually goes white when the emitted light ranges over a bigger spectrum, covering the whole visible wavelengths.
This happens to be due to blackbody radiation and Planck’s law in which higher temperatures correspond to higher-energy radiation while, at the same time, there is a shift in the emission spectrum.
Click here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
State and illustrate When’s displacement law. Give its importance.
Wien's Displacement Law states that the wavelength (λ_max) at which the intensity of radiation emitted by a black body is maximum is inversely proportional to its absolute temperature (T). In other words, as the temperature of the black body increases, the wavelength at which it emits the most radiaRead more
Wien’s Displacement Law states that the wavelength (λ_max) at which the intensity of radiation emitted by a black body is maximum is inversely proportional to its absolute temperature (T). In other words, as the temperature of the black body increases, the wavelength at which it emits the most radiation decreases.
Mathematically, Wien’s Displacement Law is given as:
λ_max = b / T
where:
– λ_max is the wavelength at which the emission intensity is maximum,
– T is the absolute temperature of the black body in Kelvin (K),
– b is Wien’s displacement constant, approximately 2.898 × 10⁻³ m·K.
Illustration:
– Object at 300 K: Using Wien’s Displacement Law, we calculate the wavelength where its radiation peaks.
λ_max = (2.898 × 10⁻³) / 300 = 9.66 μm
This is in the infrared region.
– Object at 600 K: The peak wavelength of emission is twice the object at 300 K since the temperature has an inverse proportionality to peak wavelength.
λ_max = (2.898 × 10⁻³) / 600 = 4.83 μm
It is still within the infrared, but shorter now.
Relevance of Wien’s Displacement Law:
1. Temperature Determination from Radiation:
Temperature of any object could be determined by the peak wavelength of emitted radiation under the law. It finds many applications in astrophysics, for example, to estimate the temperature of stars as a function of the color of stars. Therefore, it gives a color equivalent estimation.
2. Color of Stars
According to Wien’s law, hotter stars emit more radiation at shorter wavelengths, causing them to appear bluer, and cooler stars emit at longer wavelengths, making them appear redder.
3. Thermal Radiation Understanding:
Wien’s law helps us understand the nature of thermal radiation and how temperature affects the radiation emitted by objects, which is vital in many fields, including climatology and energy studies.
4. Temperature Measurement Applications: The law is applied in infrared thermometry, which allows objects to be measured for temperature without direct contact. This is found in industrial, medical, and scientific applications.
Click here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/