Monochromatic light of wavelength $3000 \, \mathring{A}$ is incident on a surface of area $4 \, cm^2$. If the intensity of light is $150 \, mW/m^2$, then the rate at which photons strike the target is:

$A$ bulb has a power of $60 \, mW$ and emits light of wavelength $6000 \, \mathring{A}$. Calculate the number of photons emitted by the bulb per second.

The velocity of photons is proportional to (where $\nu=$ frequency)

$A$ photon,an electron,and a uranium nucleus all have the same wavelength. The one with the most energy 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

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