Question

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?

Answer 1

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))`

Answer 2

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"`.

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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")`