The ratio of the power of a light source  $$S_1$$ to that of the light source $$S_2$$ is  2.$$S_1$$ is emitting  $$2\times10^{15}$$ 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 $$\underline {\hspace{2cm}}$$ $$\times 10^{14}.$$

We are given that the power ratio $$P_1/P_2 = 2$$, source $$S_1$$ emits $$2 \times 10^{15}$$ photons per second at 600 nm, and $$S_2$$ has wavelength 300 nm.

Since the power of a light source is given by $$P = n \cdot \frac{hc}{\lambda}$$, where $$n$$ is the number of photons emitted per second, one can write $$P_1 = n_1 \cdot \frac{hc}{\lambda_1} = 2 \times 10^{15} \cdot \frac{hc}{600 \text{ nm}}$$.

Using the ratio $$P_1/P_2 = 2$$, it follows that $$P_2 = \frac{P_1}{2} = \frac{2 \times 10^{15} \cdot hc}{2 \times 600}= \frac{10^{15} \cdot hc}{600}$$.

Moreover, since $$P_2 = n_2 \cdot \frac{hc}{\lambda_2} = n_2 \cdot \frac{hc}{300}$$, one finds $$n_2 = \frac{P_2 \cdot \lambda_2}{hc} = \frac{P_2 \times 300}{hc}$$.

Substituting the expression for $$P_2$$ yields $$n_2 = \frac{10^{15} \cdot hc \times 300}{600 \times hc} = \frac{10^{15}}{2} = 5 \times 10^{14}$$.

Therefore, the number of photons per second emitted by $$S_2$$ is $$5 \times 10^{14}$$, so the answer is $$\mathbf{5}$$.