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.
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/