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A $60 \mathrm{~W}$ source emits monochromatic light of wavelength $662.5 \mathrm{~nm}$. The number of photons emitted per second is
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Verified Answer
The correct answer is:
$2 \times 10^{20}$
Given, $P=60 \mathrm{~W}$
$\lambda=662.5 \mathrm{~nm}=6.625 \times 10^{-7} \mathrm{~m}$
Energy of 1 photon $=h v=h \frac{c}{\lambda}$
$=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6.625 \times 10^{-7}}=3 \times 10^{-19} \mathrm{~J}$
$\therefore$ Number of photons emitted per second
$\begin{aligned} & =\frac{\text { Power of source }}{\text { Energy of one photon }} \\ & =\frac{60}{3 \times 10^{-19}} \\ & =20 \times 10^{19} \\ & =2 \times 10^{20}\end{aligned}$
$\lambda=662.5 \mathrm{~nm}=6.625 \times 10^{-7} \mathrm{~m}$
Energy of 1 photon $=h v=h \frac{c}{\lambda}$
$=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6.625 \times 10^{-7}}=3 \times 10^{-19} \mathrm{~J}$
$\therefore$ Number of photons emitted per second
$\begin{aligned} & =\frac{\text { Power of source }}{\text { Energy of one photon }} \\ & =\frac{60}{3 \times 10^{-19}} \\ & =20 \times 10^{19} \\ & =2 \times 10^{20}\end{aligned}$
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