The beam of light has three wavelengths $4144 \,\mathring A$, $4972 \;\mathring A$ and $6216\; \mathring A$ with a total intensity of $3.6 \times$ $10^{-5}\,Wm ^2$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1\,cm ^2$ of a clean metallic surface of work function $2.3\,eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2\,s$.

A photon of $1.7 \times {10^{ – 13}}$ Joules is absorbed by a material under special circumstances. The correct statement is

$(i)$ In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as $E_{max} = hf – \phi _0$ (where $\phi _0$ where do is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$) what will be the maximum energy for the emitted electron ? 

$(ii)$ Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential ? 

A Laser light of wavelength $660\,nm$ is used to weld Retina detachment. If a Laser pulse of width $60\, ms$ and power $0.5\, kW$ is used the approximate number of photons in the pulse are : [Take Planck's constant $h\, = 6.62\times10^{- 34}\, Js$]

In photo electric effect

$A.$ The photocurrent is proportional to the intensity of the incident radiation.

$B.$ Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.

$C.$ Max. $K.E$ with which photoelectrons are emitted depends on the frequency of incident light.

$D.$ The emission of photoelectrons require a minimum threshold intensity of incident radiation.

$E.$ Max. K.E of the photoelectrons is independent of the frequency of the incident light.

Choose the correct answer from the options given below: