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 __________ $\times 10^{14}$.

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