Light of wavelength 5000 Å and intensity 39.8 W/m² is incident on a metal surface. If only 1% photons of incident light emit photo electrons, then what will be the number of electrons emitted/sec/unit area from the surface?
2 Answers
The energy of an individual photon is

Planck constant:
#h approx 6.63 times 10^(34) \ m^2 kg s^(1)# 
speed of light is
#c approx 3 times 10^8 \ m s^(1)# 
wavelength
#lambda = 5000 \ Å#
The number of incident photons
#N_gamma = I/E_gamma = (I lambda)/(hc)#
If only 1% photons cause an electron emission, then the number of electrons emitted per second per unit area from surface is:
Here is an alternate approach. Here we assume that the work function is overcome by the incoming light energy... and consequently get
#1.002 xx 10^(18) "photoelectrons/m"^2"/s"# .
Note that the work function was not stated, so we must assume that it is overcome by the incoming light energy...
The energy for a single photon is given by Planck:
#E_"photon" = hnu = (hc)/lambda# ,where:
#h = 6.626 xx 10^(34) "J"cdot"s"# , or#"kg"cdot"m"^2"/s"# is Planck's constant.#c = 2.998 xx 10^(8) "m/s"# is the speed of light.#lambda# is the wavelength of the incoming light in#"m"# .
The light wavelength
#5000 cancel"Å" xx (10^(10) "m")/cancel("1 Å") = 5.000 xx 10^(7) "m"#
Therefore, the energy contained in one photon of that wavelength is:
#E_"photon" = (hc)/lambda = (6.626 xx 10^(34) "J" cdot cancel"s" cdot 2.998 xx 10^8 cancel"m""/"cancel"s")/(5.000 xx 10^(7) cancel"m")#
#= 3.973 xx 10^(19) "J/photon"#
Now, the intensity of light can be treated as if it were an energy value for all the photons involved, since we retain all the units beyond
As a result, since
#(39.8 cancel"J""/m"^2"/s")/(3.973 xx 10^(19) cancel"J""/photon") = 1.002 xx 10^20 "photons/m"^2"/s"#
However, since only
#color(blue)("Number of photoelectrons") = 0.01 cdot 1.002 xx 10^20 "photoelectrons/m"^2"/s"#
#= color(blue)(1.002 xx 10^18 "photoelectrons/m"^2"/s")#