$A$ radio station broadcasts at a wavelength of $300 \ m$. If the radiating power of the transmitter is $10 \ kW$,then the number of photons emitted per second is:

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons',even in barely detectable light.
$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power,emitting radiowaves of wavelength $500\; m$.
$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $(10^{-10}\; W m^{-2})$. Take the area of the pupil to be about $0.4\; cm^2$,and the average frequency of white light to be about $6 \times 10^{14}\; Hz$.

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:

What is called as the quantum of energy? What is its energy?

An electron-positron pair is created by $\gamma$-rays. What is the minimum energy of the $\gamma$-rays required for this process in $MeV$?

Monochromatic light of wavelength $667 \, nm$ is produced by a helium-neon laser. The power emitted is $9 \, mW$. The number of photons arriving per second on average at a target irradiated by this beam is: