
Sun radiates thermal radiation with maximum intensity at the wavelength $\lambda = 0.5\mu m$ while its surface temperature is $6000\,K$. If sun cools down to a temperature where it emits only $81\% $of its present power, the maximum intensity will then be emitted at wavelength $\lambda $ (in micrometer) is equal to $[\sqrt {10} = 3.1622]$.
Answer
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Hint: The above problem is based on Wien's displacement law. This law explains the relationship between the maximum light wavelengths that a source can emit. The Wien's constant is the constant that results from multiplying the light's wavelength by its temperature.
Formula used:
The expression of Wien’s displacement law is,
${\lambda _{\max }}T = \text{Constant}$
Here, $\lambda _{\max }$ is the maximum value of the wavelength and $T$ is the temperature.
Complete step by step solution:
In the question the temperature of the sun is $T = 6000\,K$, maximum wavelength of radiation emitted $\lambda = 0.5\mu m$ and the sun cools to a temperature where it emits $81\% $ of present power. Let’s assume the temperature after cooling be $T'$ and the maximum wavelength emitted after cooling be $\lambda '$. As we know the energy emitted per unit time or power emitted by any body $\mu = e\sigma A{T'^4}$.
To find the temperature after cooling, compare the given information with the energy emitted per unit time, then we have:
$\dfrac{{0.81{\mu _1}}}{{{\mu _1}}} = \dfrac{{e\sigma A{{T'}^4}}}{{e\sigma A{{(6000)}^4}}} \\$
$\Rightarrow 0.81 \times {(6000)^4} = {{T'}^4} \\$
$\Rightarrow T' = \sqrt[4]{{1.0497 \times {{10}^{15}}}}K \\$
$\Rightarrow T' = 5692.099 \approx 5692.1\,K \\$
Apply the Wien’s displacement law to calculate the formula for temperature of the sun ${\lambda _{\max }}T = \text{Constant}$
We have the constant as $\lambda T = \lambda 'T'$ then substitute the given values, we obtain:
$0.5\mu m \times 6000\,K = \lambda ' \times 5692.1\,K \\$
$\Rightarrow \lambda ' = \dfrac{{0.5\mu m \times 6000\,K}}{{5692.1\,K}} \\$
$\Rightarrow \lambda ' = 0.527\mu m \\$
$\therefore \lambda ' \approx 0.53\mu m \\$
Therefore, the maximum wavelength emitted after cooling is approximately $0.53\mu m$.
Additional information: When a body is heated up, its thermal radiation becomes apparent. For shorter wavelengths, the radiation's maximum temperature can be observed. The relationship between the body's temperature and the thermal radiation's wavelength is inverse. The overall emissive power of the body is determined by the area under the curve between its wavelength and temperature.
Note: Before using the Wien's displacement law, it is necessary to convert the wavelength in metres and the temperature in Kelvin from degrees Celsius. For a shorter wavelength of light, the body emits the higher thermal radiation.
Formula used:
The expression of Wien’s displacement law is,
${\lambda _{\max }}T = \text{Constant}$
Here, $\lambda _{\max }$ is the maximum value of the wavelength and $T$ is the temperature.
Complete step by step solution:
In the question the temperature of the sun is $T = 6000\,K$, maximum wavelength of radiation emitted $\lambda = 0.5\mu m$ and the sun cools to a temperature where it emits $81\% $ of present power. Let’s assume the temperature after cooling be $T'$ and the maximum wavelength emitted after cooling be $\lambda '$. As we know the energy emitted per unit time or power emitted by any body $\mu = e\sigma A{T'^4}$.
To find the temperature after cooling, compare the given information with the energy emitted per unit time, then we have:
$\dfrac{{0.81{\mu _1}}}{{{\mu _1}}} = \dfrac{{e\sigma A{{T'}^4}}}{{e\sigma A{{(6000)}^4}}} \\$
$\Rightarrow 0.81 \times {(6000)^4} = {{T'}^4} \\$
$\Rightarrow T' = \sqrt[4]{{1.0497 \times {{10}^{15}}}}K \\$
$\Rightarrow T' = 5692.099 \approx 5692.1\,K \\$
Apply the Wien’s displacement law to calculate the formula for temperature of the sun ${\lambda _{\max }}T = \text{Constant}$
We have the constant as $\lambda T = \lambda 'T'$ then substitute the given values, we obtain:
$0.5\mu m \times 6000\,K = \lambda ' \times 5692.1\,K \\$
$\Rightarrow \lambda ' = \dfrac{{0.5\mu m \times 6000\,K}}{{5692.1\,K}} \\$
$\Rightarrow \lambda ' = 0.527\mu m \\$
$\therefore \lambda ' \approx 0.53\mu m \\$
Therefore, the maximum wavelength emitted after cooling is approximately $0.53\mu m$.
Additional information: When a body is heated up, its thermal radiation becomes apparent. For shorter wavelengths, the radiation's maximum temperature can be observed. The relationship between the body's temperature and the thermal radiation's wavelength is inverse. The overall emissive power of the body is determined by the area under the curve between its wavelength and temperature.
Note: Before using the Wien's displacement law, it is necessary to convert the wavelength in metres and the temperature in Kelvin from degrees Celsius. For a shorter wavelength of light, the body emits the higher thermal radiation.
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