The velocity of a photon is proportional to (where $\nu$ is frequency):

If the energy of a photon is expressed in units of $KeV$ and the wavelength in units of $\mathring{A}$,then the energy of the photon can be calculated by:

$A$ photon has a wavelength of $3 \,nm$. Its momentum and energy, respectively, will be:
$[h = 6.63 \times 10^{-34} \,Js, c = 3 \times 10^8 \,m/s]$

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

The antiparticle of an electron is:

Find the number of photons emitted per second by a $100\, W$ red light source. Assume for simplicity that the average wavelength of each photon is $694\, nm$.