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A 25 watt bulb emits monochromatic yellow light of wavelength of \(0.57 \mu \mathrm{m}\). Calculate the rate of emission of quanta per second.
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Energy emitted by the bulb \(=25 \mathrm{watt}=25 \mathrm{Js}^{-1}\)
\(\left(\because 1 \text { watt }=1 \mathrm{Js}^{-1}\right)\)
Energy of one photon \((\mathrm{E})=h v=h \frac{\mathrm{c}}{\lambda}\)
Here \(\lambda=0.57 \mu \mathrm{m}=0.57 \times 10^{-6} \mathrm{~m}\)
\(\left(\because 1 \mu \mathrm{m}=10^{-6} \mathrm{~m}\right)\)
Putting \(\mathrm{c}=3 \times 10^{-8} \mathrm{~ms}^{-1}\)
\(h=6.62 \times 10^{-34} \mathrm{Js}\), we get
\(\mathrm{E}=\frac{\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{0.57 \times 10^{-6} \mathrm{~m}}=3.48 \times 10^{-19} \mathrm{~J}\)
\(\because \quad\) No. of photons emitted per sec
\(=\frac{25 \mathrm{Js}^{-1}}{3.48 \times 10^{-19} \mathrm{~J}}=7.18 \times 10^{19}\)
\(\left(\because 1 \text { watt }=1 \mathrm{Js}^{-1}\right)\)
Energy of one photon \((\mathrm{E})=h v=h \frac{\mathrm{c}}{\lambda}\)
Here \(\lambda=0.57 \mu \mathrm{m}=0.57 \times 10^{-6} \mathrm{~m}\)
\(\left(\because 1 \mu \mathrm{m}=10^{-6} \mathrm{~m}\right)\)
Putting \(\mathrm{c}=3 \times 10^{-8} \mathrm{~ms}^{-1}\)
\(h=6.62 \times 10^{-34} \mathrm{Js}\), we get
\(\mathrm{E}=\frac{\left(6.62 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{0.57 \times 10^{-6} \mathrm{~m}}=3.48 \times 10^{-19} \mathrm{~J}\)
\(\because \quad\) No. of photons emitted per sec
\(=\frac{25 \mathrm{Js}^{-1}}{3.48 \times 10^{-19} \mathrm{~J}}=7.18 \times 10^{19}\)
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