The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\, \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3}$ $watt$ will be : $\left(\mathrm{h}=6.6 \times 10^{-34}\, \mathrm{Js}\right)$

In photoelectric effect, the electrons are ejected from metals if the incident light has a certain minimum

Light of intensity $10^{-5}\; W m ^{-2}$ falls on a sodium photo-cell of surface area $2 \;cm ^{2}$. Assuming that the top $5$ layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $ 2\; eV$. What is the implication of your answer?

A beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10 \%$ of the incident photons effectively produce photoelectron, then find current due to these electrons $………..\mu A$

[given, $hc =1240\,eV – nm , e =1.6 \times 10^{-{ }^{19}\,C }$ )

A photo sensitive material is at $9\,m$ to the left of the origin and the source of light is at $7\,m $ to the right of the origin along $x$ -axis. The photosensitive material and the source of light start from rest and move, respectively, with $8 \widehat i \, m / s$ and $4 \widehat i \, m / s$ . The ratio of intensities at $t = 0$ to $t = 3\,s$ as received by the photosensitive material is :-