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CBSE Class 11-science Answered

 A source of monochromatic electromagnetic radiation having wavelength 62nm is subjected to a cross-sectional area of 1.6 nm^2.If rate at which photons are absorbed is 1000 photons per second then calculate intensity of absorbed light in units J/m^2 sec.
Asked by anjali shrivastava | 10 Aug, 2021, 08:18: PM
answered-by-expert Expert Answer
Intensity is calculated by-
straight I equals nE over tA straight n over straight t equals IA over straight E 1000 equals fraction numerator straight I cross times 1.6 cross times 10 to the power of negative 9 end exponent cross times 10 to the power of negative 9 end exponent over denominator begin display style hc over straight lambda end style end fraction 1000 equals fraction numerator straight I cross times 1.6 cross times 10 to the power of negative 9 end exponent cross times 10 to the power of negative 9 end exponent over denominator 6.62 cross times 10 to the power of negative 34 end exponent cross times 3 cross times 10 to the power of 8 end fraction straight I equals 12.412 cross times 10 to the power of negative 16 end exponent space Jm to the power of negative 2 end exponent Sec to the power of negative 1 end exponent
Answered by Ravi | 12 Aug, 2021, 02:56: PM

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