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Question
A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) 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. (b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.
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Solution
Given :-
Wavelength of light beam, λ = 400 nm
Work function of metal plate, `phi = 2.2 "eV"`
Energy of the photon,
`E = (hc)/λ`,
Where h = Planck's constant
c = speed of light
`therefore E = 1240/400 = 3.1 "eV"`
This energy will be supplied to the electrons.
Energy lost by the electron in the first collision
= 3.1 eV × 10%
= 0.31 eV
Now, the energy of the electron after the first collision = `3.1 - 0.31 = 2.79 "eV"`
Energy lost by electron in the second collision
= 2.79 eV× 10%
= 0.279 eV
Total energy lost by the electron in two collisions
= 0.31 + 0.279 = 0.589 eV
Using Einstein's photoelectric equation, kinetic energy of the photoelectron when it comes out from the metal,
`K = E - phi - "energy lost due to collisions"`
= (3.1 − 2.2 − 0.589) eV
= 0.31 eV
(b) Similarly for the third collision, the energy lost = (2.79 - 0.279) eV × 10% = 0.2511 eV
Energy of the electron after the third collision = 2.790 - 0.2511 = 2.5389
Energy lost in the fourth collision = 2.5389 × 10%
Energy of the electron after the fourth collision = 2.5389 - 0.25389 = 2.28501
This value is very close to the work function of the metal plate. After the fifth collision, the energy of the electron becomes less than the work function of the metal.
Therefore, the electron can suffer maximum four collisions before it becomes unable to come out of the metal.
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