If the total energy transferred to a surface in time $t$ is $6.48 \times 10^5 \mathrm{~J}$, then the magnitude of the total momentum delivered to this surface for complete absorption will be :

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 percetve $(-10^{-10}\; W m ^{-2}$). 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 a source of power $4\,kW$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called

An AIR station is broadcasting the waves of wavelength $300$ metres. If the radiating power of the transmitter is $10 kW$, then the number of photons radiated per second is

Photon and electron are given same energy $({10^{ - 20}}J)$. Wavelength associated with photon and electron are ${\lambda _{Ph}}$ and ${\lambda _{el}}$ then correct statement will be

Assertion : Mass of moving photon varies inversely as the wavelength.
Reason : Energy of the particle $= mass\times(speed \,of \,light)^2$