Question

5% energy of a monochromatic light source is converted into light of wavelength 5600 $A^0$. If the power of the source be 100 W, then how many photons of light will be emitted by it per second?

Answer 1

$\lambda =5600A$

Energy of one photon is $E=h\nu =h\times \frac{c}{\lambda }$$E=6.63\times {10}^{3}4\times 3\times {10}^8/(5600\times {10}^{-10})=3.616\times {10}^{-19}J$

A$100watt$ bulb requires $100joule$ of energy per second.

Energy radiated by the bulb as visible light per second =$5$% of$100=5Joule$

Number of photons emitted per second=Energy radiated per sec/ energy of one photon

$=5/(3.616\times {10}^{-19})$

$=1.38\times {10}^{19}$