An atom absorbs a photon of wavelength $500\,nm$ and emits another photon of wavelength $600\,nm$. The net energy absorbed by the atom in this process is $n \times 10^{-4}\,eV$. The value of $n$ is ............ [Assume the atom to be stationary during the absorption and emission process] (Take $h = 6.6 \times 10^{-34}\,Js$ and $c = 3 \times 10^8\,m/s$)

Monochromatic light of wavelength $667 \, nm$ is produced by a helium-neon laser. The power emitted is $9 \, mW$. The number of photons arriving per second on average at a target irradiated by this beam is:

When an $X$-ray photon collides with a proton,its frequency:

$A$ photon,an electron,and a uranium nucleus all have the same wavelength. The one with the most energy 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 $6 \times 10^{14}\; Hz$.

Find the number of photons emitted per second by a $100\, W$ red light source. Assume for simplicity that the average wavelength of each photon is $694\, nm$.