Physics, asked by badalkushwahak1119, 1 year ago

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

Answers

Answered by harshbaweja32
78
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 

harshbaweja32: mark it brainliest
Answered by talasilavijaya
1

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}                                  

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