Question

A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency, then the number of photons emitted by the bulb in 20 seconds are
$(1eV=1.6\times {10}^{-19}J,hc=12400eV)$

• A. $2\times {10}^{18}$
• B. ${10}^{18}$
• C. ${10}^{21}$
• D. $2{\times 10}^{21}$

Answer 1

Answer: (D)

$2{\times 10}^{21}$

$\begin{array}{c}Powerofabulb\left(p\right)=40W\\ energyemittedbyabulb=power\times Time\left(s\right)=40\times 20=800J\\ Energyofphotonsemittedbyabulb=\left(\frac{80}{100}\right)\times 800=640J\\ Wavelength\left(\lambda \right)=620nm\\ Energyofaphoton=\frac{hc}{\lambda }=\frac{12400\times 1.6\times {10}^{-19}\times {10}^{-10}}{620\times {10}^{-9}}\\ =\frac{640}{3.2\times {10}^{-19}}=2\times {10}^{21}photons\end{array}$

Hence,

option $\left(D\right)$ is correct answer.