The ratio of the power of a light source S₁ to that the light source S₂ is 2. S₁ is emitting 2 × 10¹⁵ photons per second at 600 nm. If the wavelength of the source S₂ is 300 nm, then the number of photons per second emitted by S₂ is_____×10¹⁴.

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