Specific heat of water is $4.2 \, J/g^{\circ}C$. If light of frequency $3 \times 10^9 \, Hz$ is used to heat $400 \, g$ of water from $20^{\circ}C$ to $40^{\circ}C$,the number of photons needed 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 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 $6 \times 10^{14}\; Hz$.

$(a)$ When monochromatic light is incident on a surface separating two media,the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
$(b)$ When light travels from a rarer to a denser medium,the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave?
$(c)$ In the wave picture of light,intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light?

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

In the Compton scattering process,the incident $X$-radiation is scattered at an angle of $60^{\circ}$. The wavelength of the scattered radiation is $0.22 \ \text{Å}$. The wavelength of the incident $X$-radiation in $\text{Å}$ units is:

An important spectral emission line has a wavelength of $21 \ cm$. The corresponding photon energy is $(h = 6.62 \times 10^{-34} \ J \cdot s; \ c = 3 \times 10^8 \ m/s)$.