The frequency of a photon having energy $100 \ eV$ is $(h = 6.6 \times 10^{-34} \ J \cdot s)$.

The number of photons of wavelength $540 \ nm$ emitted per second by an electric bulb of power $100 \ W$ is (taking $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$):

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

If the wavelength of an $X-ray$ is $0.010 \ \mathring A$,what is its momentum?

The momentum of a photon with frequency $1.5 \times 10^{13} \ Hz$ is .......