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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    A 10 kW transmitter emits radio waves of wavelength 500 m. The number of photons emitted per second by the transmitter is of the order of   JEE Main Online Paper (Held On 12 May 2012)

    A) 1037                                       

    B)                        1031

    C)                        1025                                       

    D)                        1043

    Correct Answer: B

    Solution :

                    Power \[=\frac{nhc}{\lambda }\](where, n = no. of photons per second)\[\Rightarrow \]\[n=\frac{10\times {{10}^{3}}\times 500}{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}\simeq {{10}^{31}}\]


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