If the energy of a photon is $E$ and its momentum is $p$,given by the relations $E = h\nu$ and $p = h/\lambda$,what is the velocity of the photon?

The rest energy of an electron is

$A$ gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is $0.5\, MeV$ and the total kinetic energy $(K.E.)$ of the electron-positron pair is $0.78\, MeV$,then the energy of the gamma ray photon must be.......$MeV$.

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

$A$ $10\, kW$ transmitter emits radio waves of wavelength $500\, m$. The number of photons emitted per second by the transmitter is of the order of

The frequency of a photon having energy $66 \ eV$ is: