The ratio of the power of a light source $S_1$ to that the light source $S_2$ is $2$. $S_1$ is emitting $2 imes10^{15}$ 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 ...... $ imes10^{14}$.

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 $2	imes10^{15}$ 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 ...... $	imes10^{14}$.

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Power $=N\frac{hc}\lambda$, so $\frac{P_1}{P_2}=\frac{N_1\lambda_2}{N_2\lambda_1}=2\Rightarrow N_2=\frac{N_1\lambda_2}{2\lambda_1}=\frac{2	imes10^{15}	imes300}{2	imes600}=5	imes10^{14}$. Answer: 5.

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

Explanation

Reviewed by Dr. Abubacker

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