# 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?

Jun 18, 2018

The energy of an individual photon is ${E}_{\gamma} = \frac{h c}{\lambda}$, where:

• Planck constant: $h \approx 6.63 \times {10}^{- 34} \setminus {m}^{2} k g {s}^{- 1}$

• speed of light is $c \approx 3 \times {10}^{8} \setminus m {s}^{- 1}$

• wavelength lambda = 5000 \ Å

The number of incident photons $N$ per second per square metre is therefore:

• ${N}_{\gamma} = \frac{I}{E} _ \gamma = \frac{I \lambda}{h c}$

If only 1% photons cause an electron emission, then the number of electrons emitted per second per unit area from surface is:

${N}_{e} = 0.01 {N}_{\gamma} = 0.01 \cdot \frac{39.8 \cdot 5000 \times {10}^{- 10}}{3 \times {10}^{8} \cdot 6.63 \times {10}^{- 34}}$

$\boldsymbol{\therefore {N}_{e} \approx {10}^{18} q \quad {m}^{- 2} {s}^{- 1}}$

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

$1.002 \times {10}^{18} \text{photoelectrons/m"^2"/s}$.

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:

${E}_{\text{photon}} = h \nu = \frac{h c}{\lambda}$,

where:

• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$, or $\text{kg"cdot"m"^2"/s}$ is Planck's constant.
• $c = 2.998 \times {10}^{8} \text{m/s}$ is the speed of light.
• $\lambda$ is the wavelength of the incoming light in $\text{m}$.

The light wavelength $5000$ angstroms should be represented in $\text{m}$, as Planck's constant has $\text{m}$.

$5000 \cancel{\text{Å" xx (10^(-10) "m")/cancel("1 Å") = 5.000 xx 10^(-7) "m}}$

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")

$= 3.973 \times {10}^{- 19} \text{J/photon}$

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 $\text{J}$.

As a result, since $\text{1 W}$ $=$ $\text{1 J/s}$, we have that $\text{1 W/m"^2 = "1 J/m"^2"/s}$, 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"

$= \textcolor{b l u e}{1.002 \times {10}^{18} \text{photoelectrons/m"^2"/s}}$