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GUJCET · Physics · Dual Nature of Radiation and Matter
The number of photons emitted per second by a bulb of 66 W power emitting waves of wavelength 600 nm is _________ . \(\left(h=6.6 \times 10^{-34} \mathrm{Js}\right)\)
- A \(2 \times 10^{22}\)
- B \(2 \times 10^{19}\)
- C \(2 \times 10^{21}\)
- D \(2 \times 10^{20}\)
Answer & Solution
Correct Answer
(D) \(2 \times 10^{20}\)
Step-by-step Solution
Detailed explanation
(D) \(2 \times 10^{20}\)
Energy of Photon
\(
\mathrm{E}=\frac{n h c}{\lambda}
\)
\(
\begin{array}{ll}
\therefore & \mathrm{P} t=\frac{n h c}{\lambda} \quad\left(\because \mathrm{P}=\frac{\mathrm{E}}{t}\right) \\
\therefore & n=\frac{\mathrm{P} t \lambda}{h c} \\
\therefore & n=\frac{66 \times 1 \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^{8}} \\
\therefore & n=20 \times 10^{19} \\
\therefore & n=2 \times 10^{20}
\end{array}
\)
Energy of Photon
\(
\mathrm{E}=\frac{n h c}{\lambda}
\)
\(
\begin{array}{ll}
\therefore & \mathrm{P} t=\frac{n h c}{\lambda} \quad\left(\because \mathrm{P}=\frac{\mathrm{E}}{t}\right) \\
\therefore & n=\frac{\mathrm{P} t \lambda}{h c} \\
\therefore & n=\frac{66 \times 1 \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^{8}} \\
\therefore & n=20 \times 10^{19} \\
\therefore & n=2 \times 10^{20}
\end{array}
\)
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