Hey, do you want to learn about the Einstein equation of the photoelectric effect? If yes. Then keep reading.

## Einstein's photoelectric equation

- Einstein explained the photoelectric effect using the quantum nature of radiation.
- The emission of a photoelectron is a result of the interaction of one photon with a loosely bound electron in which the photon is completely absorbed by the electron.
- Some part of incident energy equal to work function is used to remove an electron from metal and remaining is given to electron as its kinetic energy.

$\mathrm{hv}=\mathrm{W}+\mathrm{E}_{\max }=\mathrm{W}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$

$=h v_{0}+\frac{1}{2} m v_{\max }^{2}$

$\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{h}\left(\mathrm{v}-\mathrm{v}_{0}\right)$

$=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\mathrm{eV}_{\mathrm{S}} \ldots(1)$

equation 1 is called Einstein photoelectric equation. maximum velocity of the emitted electron

$v_{\max }=\sqrt{\frac{2 h\left(v-v_{0}\right)}{m}}=\sqrt{\frac{2 h c\left(\lambda-\lambda_{0}\right)}{m \lambda \lambda_{0}}}$ $=\sqrt{\frac{2 \mathrm{eV}_{\mathrm{S}}}{\mathrm{m}}}$

The stopping potential

$V_{S}=\frac{h\left(v-v_{0}\right)}{e}=\frac{h c\left(\lambda_{0}-\lambda\right)}{e \lambda \lambda_{0}}$ - The Einstein's photoelectric equation is in accordance with the conservation of energy. Here light energy is converted into electric energy.
- The equation explains the laws of photoelectric emission.

(a) The increase in intensity increases the number of photons with the same energy hv. So the number of photoelectrons will proportionally increase.

(b) If $vthen KE will become negative which is not possible so in this condition photoemission is not possible.

(c) If $v>v_{0}$ Then $\mathrm{KE} \propto\left(v-v_{0}\right)$ so maximum kinetic energy or stopping potential increases linearly with frequency of incident radiation.

## Important points

- In the photoelectric effect, all photoelectrons do not have the same kinetic energy. Their KE ranges from zero to $E_{\max }$ which depends on the frequency of incident radiation and nature of cathode.
- The photoelectric effect takes place only when photons strikebound electrons because for free electrons energy and momentum conservations do not hold together.
- Cesium is the best photo-sensitive material.
- The efficiency of a photoemission

$\eta=\frac{\text { Number of photoelectrons emitted per unit area per unit time }}{\text { Number of photons incident per unit area per unit time }}$

$=\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{p}}}$ $\eta=\frac{\text { Intensity of emitted electrons }}{\text { Intensity of incident radiation }}$

$=\frac{I_{e}}{I_{p}}$ Therefore $\eta=\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{p}}}=\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{I}_{\mathrm{p}}}$

so, that's all from this article. I hope you get the idea about the Einstein equation of the photoelectric effect. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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