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(C) The power $P$ of a source is given by $P = n \cdot E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\

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Infrared source

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(C) The power $P$ of a source is given by $P = n \cdot E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}$.Therefore,the number of photons emitted per second is $n = \frac{P \cdot \lambda}{hc}$.Since the power $P$ and the constants $h$ and $c$ are the same for all three sources,the number of photons $n$ is directly proportional to the wavelength $\lambda$ $(n \propto \lambda)$.Comparing the wavelengths: $\lambda_{UV} \approx 400 \ nm$,$\lambda_{Visible} \approx 550 \ nm$,and $\lambda_{IR} \approx 700 \ nm$.Since $\lambda_{IR} > \lambda_{Visible} > \lambda_{UV}$,the number of photons emitted per second is maximum for the infrared source.

If sources of ultraviolet $(\lambda \approx 400 \ nm)$,visible $(\lambda \sim 550 \ nm)$,and infrared $(\lambda \sim 700 \ nm)$ radiation each have a power rating of $100 \ W$,then the number of photons emitted per unit area per unit time is maximum for:

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