Question

All of its output energy is absorbed in a detector that measures a total energy of 0.54 JJ over a period of 33 ss . How many photons per second are being emitted by the laser? Express your answer using two significant figures.

Answers

  1. Answer:

    7.84 x 10¹⁶ photons/s

    Explanation:

    Given,

    Total energy = 0.54 J

    Period = 33 s

    Photons per second = ?

    Wavelength of the Photon = 989 nm

    Energy of  single photon:

    E =\frac{h c}{\lambda}

    =\frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8}}{989 \times 10^{-9}}

    =2.01 \times 10^{-19} \mathrm{J}

    Total energy

    Time period =33 \mathrm{s}

    Energy absorbed per sec =\frac{0.52}{33 }=0.01575 \mathrm{J} / \mathrm{s}

    Energy of a photon =2.01 \times 10^{-19} \mathrm{J} / photon

    Number of photons emitted per second is

    \frac{0.01575}{2.01 \times 10^{-19}} \mathrm{J} / \text { photon }}

    Number of photons = 7.84 x 10¹⁶ photons/s

Leave a Comment