CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The number of photons emitted per second by a 60 W source of monochromatic light of wavelength 663 nm is \[(h=6.63\times {{10}^{-34}}Js)\]

    A) \[4\times {{10}^{-20}}\]               

    B) \[1.5\times {{10}^{20}}\]

    C) \[3\times {{10}^{-20}}\]         

    D)        \[2\times {{10}^{20}}\]

    E) \[1\times {{10}^{-20}}\]

    Correct Answer: D

    Solution :

    Energy, \[E=\frac{nhc}{\lambda }\] \[\Rightarrow \]\[60\times 1\,Js=\frac{n\times 6.63\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m}{663\times {{10}^{-9}}m}\]                                 \[\left[ \because power=\frac{energy}{time} \right]\] \[\therefore \]  \[n=\frac{60\times 1\times 663\times {{10}^{-9}}}{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}\]                 \[=2\times {{10}^{20}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner