KCET · Physics · Dual Nature of Matter
A \(60 \mathrm{~W}\) source emits monochromatic light of wavelength \(662.5 \mathrm{~nm}\). The number of photons emitted per second is
- A \(5 \times 10^{17}\)
- B \(2 \times 10^{20}\)
- C \(5 \times 10^{26}\)
- D \(2 \times 10^{29}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{20}\)
Step-by-step Solution
Detailed explanation
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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