Respuesta :
Answer:
a) wavelength, [tex]\lambda _{maximum} = 3.214\times 10^{-7} m[/tex]
b) Surface temperature, T = 9017.89 K
Given:
Total Intensity, P = 375 MWm = 375[tex]\times 10^{6} Wm[/tex]
Here, total intensity is the power per unit area
Solution:
a) Using Wien's displacement law which gives the inverse relation between temperature and wavelength for black body radiation and is given by:
[tex]\lambda _{maximum} = \frac{k}{T}[/tex] (1)
where
k = proportionality constant = 2898 [tex]\micro m.K[/tex]
[tex]\lambda _{maximum}[/tex] = maximum wavelength
T = Temperature in kelvin
From eqn (1):
[tex]\lambda _{maximum} = \frac{2898\times 10^{-6}}{T}[/tex] {2}
b) Temperature, T at the surface is given by Stefan-Boltzman law:
P = [tex]\sigma T^{4}[/tex] (3)
where
[tex]\sigma = Stefan-Boltzman = 5.670373\times 10^{-8} W/m^{2}K^{4}[/tex]
Using eqn (3):
[tex]T^{4} = \frac{P}{\sigma }[/tex]
[tex]T^{4} = \frac{375\times 10^{6}}{5.670373\times 10^{-8}} = 6.61332\times 10^{15}[/tex]
T = 9017.89 K
Now, substituting the value of T = 9017.89 K in eqn (2):
[tex]\lambda _{maximum} = \frac{2898\times 10^{-6}}{9017.89} = 3.214\times 10^{-7} m[/tex]
[tex]\lambda _{maximum} = 3.214\times 10^{-7} m[/tex]
