Question 11 of 191
The ratio of the power of a light source to that the light source is is emitting photons per second at 600 nm . If the wavelength of the source is 300 nm , then the number of photons per second emitted by is .
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Answer: 5
Since power emitting by a source is given as
$\begin{aligned}
& =\frac{\text { Total energy emitted }}{\text { time }} \\
& =\frac{\left(E_1 \text { photon }\right) \times \text { Number of photons }(N)}{t} \\
& P_1=\left(E_1\right) n
\end{aligned}$
$\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\left(\mathrm{E}_1\right) \mathrm{n}_1}{\left(\mathrm{E}_2\right) \mathrm{n}_2}=\frac{\left(\frac{\mathrm{hC}}{\lambda_1}\right) \mathrm{n}_1}{\left(\frac{\mathrm{hC}}{\lambda_2}\right) \mathrm{n}_2} \\
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \frac{\mathrm{n}_1}{\mathrm{n}_2}
\end{aligned}$
Substituting the given values
$\begin{aligned}
& 2=\left(\frac{300}{600}\right) \times \frac{2 \times 10^{15}}{\mathrm{n}_2} \\
& \mathrm{n}_2=\frac{1}{2} \times 10^{15}=5 \times 10^{14} \text { Photon } / \mathrm{sec}
\end{aligned}$
$\begin{aligned}
& =\frac{\text { Total energy emitted }}{\text { time }} \\
& =\frac{\left(E_1 \text { photon }\right) \times \text { Number of photons }(N)}{t} \\
& P_1=\left(E_1\right) n
\end{aligned}$
$\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\left(\mathrm{E}_1\right) \mathrm{n}_1}{\left(\mathrm{E}_2\right) \mathrm{n}_2}=\frac{\left(\frac{\mathrm{hC}}{\lambda_1}\right) \mathrm{n}_1}{\left(\frac{\mathrm{hC}}{\lambda_2}\right) \mathrm{n}_2} \\
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \frac{\mathrm{n}_1}{\mathrm{n}_2}
\end{aligned}$
Substituting the given values
$\begin{aligned}
& 2=\left(\frac{300}{600}\right) \times \frac{2 \times 10^{15}}{\mathrm{n}_2} \\
& \mathrm{n}_2=\frac{1}{2} \times 10^{15}=5 \times 10^{14} \text { Photon } / \mathrm{sec}
\end{aligned}$