The ratio of the power of a light source S_1 | vedclass

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(C) The power $P$ emitted by a source is given by the total energy emitted per unit time.$P = \frac{N \cdot E_{photon}}{t} = n \cdot \frac{hc}{\lambda

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(C) The power $P$ emitted by a source is given by the total energy emitted per unit time.$P = \frac{N \cdot E_{photon}}{t} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons per second.Given the ratio of powers $\frac{P_1}{P_2} = 2$.Using the formula $\frac{P_1}{P_2} = \frac{n_1 \cdot (hc / \lambda_1)}{n_2 \cdot (hc / \lambda_2)} = \frac{n_1 \lambda_2}{n_2 \lambda_1}$.Substituting the given values: $2 = \frac{(2 	imes 10^{15}) 	imes 300}{n_2 	imes 600}$.$2 = \frac{2 	imes 10^{15}}{2 \cdot n_2}$.$2 = \frac{10^{15}}{n_2}$.$n_2 = \frac{10^{15}}{2} = 0.5 	imes 10^{15} = 5 	imes 10^{14}$ photons per second.

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 $2 \times 10^{15}$ 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 . . . . . . $\times 10^{14}$.

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