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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 some collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the minimum number of collisions the electron can suffer before it is unable to come out of the metal.

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answer is 4.

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Detailed Solution

Energy of photon, E $= \frac{h c}{λ} = \frac{1240}{400} = 3.1 eV$

Energy after first collision = 0.9E

Energy after the second collision is = 0.9 × 0.9E = 0.81E

Energy after n^th collision will be (0.9)ⁿE.

Electron will not be able to come out if (0.9)ⁿE becomes less than 2.2 eV

So n = 4 will be minimum required collision to stop it from ejection.

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