What is the energy and momentum of a photon having frequency $v$?

$A$ bulb lamp emits light of mean wavelength of $4500 \mathring A$. The lamp is rated at $150 \text{ W}$ and $8 \%$ of the energy appears as emitted light. How many photons are emitted by the lamp per second?

$A$ photodiode sensor is used to measure the output of a $300 W$ lamp kept $10 m$ away. The sensor has an opening of $2 cm$ in diameter. How many photons enter the sensor if the wavelength of the light is $660 nm$ and the exposure time is $100 ms$? (Assume that all the energy of the lamp is given off as light and $h = 6.6 \times 10^{-34} J s$)

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

$A$ photon of wavelength $\lambda$ is scattered from an electron,which was at rest. The wavelength shift $\Delta \lambda$ is three times of $\lambda$ and the angle of scattering $\theta$ is $60^o$. The angle at which the electron recoiled is $\phi$. The value of $\tan \phi$ is: (electron speed is much smaller than the speed of light)

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