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A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. Under the same assumptions find the maximum number of collisions the electron should suffer before it becomes unable to come out of the metal.

A light beam of wavelength 400 nm……

Solution.

$\displaystyle E(\text{in eV})=\frac{12375}{\lambda (in\text{ angstrom})}$

$\displaystyle \Rightarrow E=\frac{12375}{4000}=3.1eV$

Energy of electron after first collision

E₁ = (90 % of E) = 2.79 eV

Energy of electron after second collision

E₂ = (90 % of E₁) = 2.51 eV

Hence, kinetic energy of the electron as it comes out of out of the metal surface

KE = (2.51 – 2.2) eV = 0.31 eV

Now

Energy of electron after third collision

E₃ = (90 % of E₂) = 2.26 eV

Energy of electron after forth collision

E₄ = (90 % of E₃) = 2.03 eV < Φ₀

After forth collision, the energy of electron is less than the work function of the metal, so the electron will not be able to come out of the metal.