The number of photons per second on an average emitted by a source of monochromatic light of wavelength $600 \, nm$,when it delivers a power of $3.3 \times 10^{-3} \, W$,is: $(h = 6.6 \times 10^{-34} \, Js)$

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 $6 \times 10^{14}\; Hz$.

$A$ photon of wavelength $4400 \, \mathring{A}$ is passing through vacuum. The effective mass and momentum of the photon are respectively

The radiant energy from the Sun incident on the Earth's surface is $2 \ cal/cm^2 \cdot min$. If the average wavelength of solar radiation is $5500 \ \mathring A$,then the number of photons incident per minute on a surface area of $1 \ cm^2$ on the Earth's surface is ............ $(h = 6.6 \times 10^{-34} \ J \cdot s, 1 \ cal = 4.2 \ J)$

The momentum of a photon in an $X-$ray beam of $10^{-10} \ m$ wavelength is

An electron and a photon have the same wavelength of $10^{-9} \, m$. If $E$ is the energy of the photon and $p$ is the momentum of the electron,then the magnitude of $E / p$ (in $SI$ unit) is