Question 840: Thermodynamics & KTG

Question

Assume that for solar radiation, surface temperature of the sun is $6000 \mathrm{~K}$. If Wien's constant 'b' is $2.897 	imes 10^{-3} \mathrm{~mK}$, the value of maximum wavelength will be

Accepted Answer

To find the value of the maximum wavelength of solar radiation, we can use Wien's Displacement Law. Wien's Displacement Law states that the product of the wavelength at which the emission of a black body radiation spectrum is maximized ($ \lambda_{	ext{max}} $) and the absolute temperature (T) of the black body is a constant. The law is mathematically represented as:
$ \lambda_{	ext{max}} 	imes T = b $
Where:
$ \lambda_{	ext{max}} $ is the wavelength corresponding to the peak emission
T is the absolute temperature
b is Wien's constant
Given:
$ T = 6000 \mathrm{~K} $
$ b = 2.897 	imes 10^{-3} \mathrm{~mK} $
Rearranging the formula to solve for $ \lambda_{	ext{max}} $:
$ \lambda_{	ext{max}} = \frac{b}{T} $
Substitute the given values into the equation:
$ \lambda_{	ext{max}} = \frac{2.897 	imes 10^{-3} \mathrm{~mK}}{6000 \mathrm{~K}} $
Calculate the value:
$ \lambda_{	ext{max}} = \frac{2.897 	imes 10^{-3}}{6000} \mathrm{~m} $
$ \lambda_{	ext{max}} = 4.828 	imes 10^{-7} \mathrm{~m} $
To convert meters to angstroms ($ \mathop A\limits^o $), note that:
$ 1~	ext{meter} = 10^{10} \mathop A\limits^o $
Therefore:
$ \lambda_{	ext{max}} = 4.828 	imes 10^{-7} \mathrm{~m} 	imes 10^{10} \mathop A\limits^o $
$ \lambda_{	ext{max}} = 4828 \mathop A\limits^o $
Therefore, the value of the maximum wavelength is $ 4828 \mathop A\limits^o $, and hence, the correct answer is:
Option A: 4828$ \mathop A\limits^o $

Answer

4828$\mathop A\limits^o $

Answer

3648$\mathop A\limits^o $

Answer

6400$\mathop A\limits^o $

Answer

5890$\mathop A\limits^o $

Thermodynamics & KTG

Assume that for solar radiation, surface temperature of the sun is 6000 \mathrm{~K}. If Wien's constant 'b' is 2.897 \times 10^{-3} \mathrm{~mK}, the value of maximum wavelength will be

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