If $n_r$ and $n_b$ are the number of photons emitted per second by a red bulb and a blue bulb of equal power respectively,then:

Light of wavelength $5000 \, \mathring{A}$ falls on a sensitive surface. If the surface has received $10^{-7} \, \text{J}$ of energy,then the number of photons falling on the surface will be:

Energy of a photon whose frequency is $10^{12} \text{ MHz}$ is
[ Planck's constant,$h = 6.63 \times 10^{-34} \text{ Js}, e = 1.6 \times 10^{-19} \text{ C}$ ]

$10^{20}$ photons of wavelength $660 \ nm$ are emitted per second from a lamp. The wattage of the lamp is (Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$). (in $W$)

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

Which of the following relations is correct for a photon?