Question

A bulb of 40w is producing a light f wavelength 620 nm with80% of efficiency then the no. of photons emitted by the bulb in 20 seconds is

Answer 1

Power of a bulb (P) = 40 W

Energy emitted by a bulb  = Power x Time(s) = 40 x 20 = 800 J

Energy of photons emitted by a bulb = (80/100) x 800 = 640 J


Wavelength of light (λ) = 620 nm
​Energy of a photon  = hcλ =12400 ×1.6×10−19×10−10 620 × 10−9 =3.2 ×10−19 J

Number of photons emitted = Total energy emittedEnergy of a photon =640 3.2×10−19                                   = 2 × 1021 photons 

Answer 2

Answer:

Number of photons emitted by the bulb with $80\%$ of efficiency is $2 \times 10^{21}$  

Explanation:

Given power of a bulb, $P = 40 W$

Wavelength of light, $\lambda = 620 nm=620\times 10^{-9} m$

Efficiency of bulb, $e=80\%$

Time, $t=20s$

Energy emitted by a bulb  = Power x Time

                                           $= 40 \times 20 = 800 J$

For $80\%$ of efficiency, energy emitted by a bulb

                                          $= \frac{80}{100} \times 800 = 640 J$

​Energy of a photon,  $E= \frac{hc}{\lambda}$

where Planck's constant, $h=6.63\times 10^{-34} J-s$

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

                              $\therefore E= \frac{6.63\times 10^{-34} \times 3\times 10^{8} }{620\times 10^{-9}}$

                                     $=\frac{6.63\times 10^{-18} \times 3 }{62}=3.2 \times 10^{-19} J$

Number of photons emitted by the bulb,

                                $N=\frac{ energy \ emitted }{Energy\ of \ a\ photon}$

                                     $=\frac{ 640 }{3.2\times 10^{-19} }$

                                     $=\frac{ 64\times 10^{20} }{3.2 }= 2 \times 10^{21}$

Number of photons emitted by the bulb is  $2 \times 10^{21}$