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

Vedclass · Accepted Answer

$0.57$

Vedclass · Answer

(A) The power $P$ of a source is given by $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.Thus,the number of photons emitted per second is $n = \frac{P \lambda}{hc}$.Given $P_{UV} = P_{IR} = 130 \ W$,$\lambda_{UV} = 400 \ nm$,and $\lambda_{IR} = 700 \ nm$.The ratio of the number of photons emitted per second is:$\frac{n_{UV}}{n_{IR}} = \frac{P_{UV} \lambda_{UV} / hc}{P_{IR} \lambda_{IR} / hc} = \frac{\lambda_{UV}}{\lambda_{IR}}$.Substituting the values:$\frac{n_{UV}}{n_{IR}} = \frac{400 \ nm}{700 \ nm} = \frac{4}{7} \approx 0.57$.

An ultraviolet bulb emits light of wavelength $400 \ nm$ and an infrared bulb emits light of wavelength $700 \ nm$. If both are rated at $130 \ W$,what is the ratio of the number of photons emitted per second by the $UV$ and $IR$ sources?

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