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CBSE Class 11-science Answered

A furnace emits radiation at 2000 K. Treating it as black body radiation, calculate Wave length at which emission is maximum and the corresponding radiant flux density
Asked by vikranthpoluparthi1 | 04 May, 2021, 12:30: PM
answered-by-expert Expert Answer
c1 = 0.374 x 10 -15 W/m2 
M a x i m u m space w a v e l e n g t h space left parenthesis lambda subscript m a x end subscript right parenthesis

B y space W e i n apostrophe s space l a w comma space
lambda subscript m a x end subscript space T space equals space 2.9 cross times 10 to the power of negative 3 end exponent space m K space
lambda subscript m a x end subscript space equals space fraction numerator 2.9 cross times 10 to the power of negative 3 end exponent over denominator T end fraction space equals space fraction numerator 2.9 cross times 10 to the power of negative 3 end exponent over denominator 2000 end fraction
equals space 1.45 cross times 10 to the power of negative 6 end exponent space m space
lambda subscript m a x end subscript space equals space 1.45 space mu

E subscript b lambda end subscript space equals space fraction numerator c subscript 1 lambda subscript m a x end subscript to the power of negative 5 end exponent over denominator e open parentheses blank to the power of begin display style fraction numerator c subscript 2 over denominator lambda subscript m a x end subscript T end fraction end style end exponent close parentheses subscript negative 1 end subscript end fraction space equals space fraction numerator 0.374 space cross times 10 to the power of negative 15 end exponent cross times open parentheses 1.45 cross times 10 to the power of negative 6 end exponent close parentheses to the power of negative 5 end exponent over denominator open parentheses begin display style fraction numerator 14.4 space cross times 10 to the power of negative 3 end exponent over denominator 1.45 cross times 10 to the power of negative 6 end exponent cross times 2000 end fraction end style close parentheses subscript negative 1 end subscript end fraction space equals space 4.09 cross times 10 to the power of 11 W divided by m squared 
 
Thus, corresponding radiant flux density = 4.09 x 1011 W/m2 
Answered by Shiwani Sawant | 04 May, 2021, 03:01: PM
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