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
Jun 18, 2018

The energy of an individual photon is #E_gamma = (h c)/lambda#, where:

  • 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# per second per square metre is therefore:

  • #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:

#N_e = 0.01 N_gamma = 0.01 * (39.8* 5000 times 10^(-10))/(3 times 10^8 * 6.63 times 10^(-34))#

#bb( :. N_e approx 10^(18) qquad m^(-2) s^(-1))#

Jun 18, 2018

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# angstroms should be represented in #"m"#, as Planck's constant has #"m"#.

#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 #"J"#.

As a result, since #"1 W"# #=# #"1 J/s"#, we have that #"1 W/m"^2 = "1 J/m"^2"/s"#, and:

#(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 #1%# of the photons cause photoemission, the number of photoelectrons is only #1%# of the number of incoming photons:

#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")#