If $c$ is the velocity of light in free space,the correct statements about photon among the following are:
$A$. The energy of a photon is $E=h v$.
$B$. The velocity of a photon is $c$.
$C$. The momentum of a photon,$p=\frac{h v}{c}$.
$D$. In a photon-electron collision,both total energy and total momentum are conserved.
$E$. Photon possesses positive charge.
Choose the correct answer from the options given below:

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

Monochromatic light of frequency $6 \times 10^{14} \text{ Hz}$ is produced by a laser. The power emitted is $4 \times 10^{-3} \text{ W}$. How many photons per second on an average are emitted by the source? $[h = 6.63 \times 10^{-34} \text{ Js}]$

$A$ particle which has zero rest mass and non-zero energy and momentum must travel with a speed

The curve drawn between the velocity and frequency of a photon in a vacuum will be a

The ratio of the power of a light source $S_1$ to that of the light source $S_2$ is $2$. $S_1$ is emitting $2 \times 10^{15}$ photons per second at $600 \ nm$. If the wavelength of the source $S_2$ is $300 \ nm$,then the number of photons per second emitted by $S_2$ is . . . . . . $\times 10^{14}$.