Question

Monochromatic light of wavelength 3000 A° is incident normally on a surface of area 4 cm². If the intensity of light is 150 mW/m², find the number of photons being incident on this surface in one second.

Answer 1

Answer:

$9.05\times 10^{15}.$

Explanation:

Given,

$\textrm{Wavelength of light}, \lambda\ =\ 3000\ A^o$

                                                 $=\ 3000\times 10^{-10}\ m$

                                                 $=\ 3\times 10^{-7}\ m$

$\textrm{area of the surface},\ A\ =\ 4\ cm^2$

                                      $=\ 4\times 10^{-2}\ m$

$\textrm{intensity of light},\ I\ =\ 150\ mW/m^2$

                                $=\ 150\times 10^{-3}\ W/m^2$

                               $=0.15\ W/m^2$

Hence, energy of the radiation falling on the surface in 1 second is given by

                                  $E\ =\ intensity\times area\times time$

                                        $=\ 0.15\times 4\times 10^{-2}\times 1 J$

                                      $=\ 0.6\times 10^{-2}\ J$

And energy of radiation can be also determined by

                             $E\ =\ \dfrac{n.h.c}{\lambda}$

            where, n = number of photon

                         $h\ =\textrm{\ planck's\ constant}\ =\ 6.625\times 10^{-34}$

                        $c\ =\textrm{\ speed of light\ }=\ 3\times 10^{8} m/s$

So,

$\ 0.6\times 10^{-2}\ =\ \dfrac{n.6.625\times 10^{-34}\times 3\times 10^8}{3\times 10^{-7}}$

$=>\ n\ =\ 9.05\times 10^{15}$

Hence, the total number of photon will be $9.05\times 10^{15}.$