Question 245: Dual Nature of Matter and Radiation

Question

The ratio of the power of a light source $S_1$ to that the light source $S_2$ is $2 . S_1$ is emitting $2 	imes 10^{15}$ photons per second at 600 nm . If the wavelength of the source $S_2$ is 300 nm , then the number of photons per second emitted by $S_2$ is __________ $	imes 10^{14}$.

Accepted Answer

$$\begin{aligned}
&	ext { Since power emitting by a source is given as }\
&\begin{aligned}
& =\frac{	ext { Total energy emitted }}{	ext { time }} \
& =\frac{\left(E_1 	ext { photon }ight) 	imes 	ext { Number of photons }(N)}{t} \
& P_1=\left(E_1ight) n
\end{aligned}
\end{aligned}$$
$$\begin{aligned}
&\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\left(\mathrm{E}_1ight) \mathrm{n}_1}{\left(\mathrm{E}_2ight) \mathrm{n}_2}=\frac{\left(\frac{\mathrm{hC}}{\lambda_1}ight) \mathrm{n}_1}{\left(\frac{\mathrm{hC}}{\lambda_2}ight) \mathrm{n}_2} \
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\lambda_2}{\lambda_1}ight) \frac{\mathrm{n}_1}{\mathrm{n}_2}
\end{aligned}\
&	ext { Substituting the given values }\
&\begin{aligned}
& 2=\left(\frac{300}{600}ight) 	imes \frac{2 	imes 10^{15}}{\mathrm{n}_2} \
& \mathrm{n}_2=\frac{1}{2} 	imes 10^{15}=5 	imes 10^{14} 	ext { Photon } / \mathrm{sec}
\end{aligned}
\end{aligned}$$

Dual Nature of Matter and Radiation

The ratio of the power of a light source S_1 to that the light source S_2 is 2 . S_1 is emitting 2 \times 10^{15} photons per second at 600 nm . If the wavelength of the source S_2 is 300 nm , then the number of photons per second emitted by S_2 is __________ \times 10^{14}.

Choose the correct answer: