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The photoelectric work function of potassium is 2.3eV. If light that has a wavelength 190nm of falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

Short Answer

Expert verified
  1. The stopping potential is 4.24eV.
  2. The kinetic energy of the most energetic electrons ejected is 4.24eV.
  3. The speed of theses electrons is 1.22*106m/s.

Step by step solution

01

Formula for maximum kinetic energy

Kmax=12mvmax2=hf-Ï• (1)

whereKmax=eV0

⇒eV0=hf-ϕ=hcλ-ϕ

02

Calculate the stopping potential

Given: e=1.6*10-19C

λ=190nm=190*10-19Ch=6.626*10-34Jsϕ=2.3eV

Substitute the given in equation (2),

V0=1e(6.626*10-34Js*3*108m/s190*10-9m)=6.54eV-2.3eV=4.24eV

03

Calculate the kinetic energy

Substitute the values in equation (1),

Kmax=6.626*10-34Js*3*108Js190*10-9m-2.3eV=6.54eV-2.3eV=4.24eV

04

Calculate the speed of electrons

From equation (1), the speed of electrons can be given as:

v=2m(hcλ-ϕ)=29.1*10-31kg(6.26*10-34Js*3*108Js190*10-9m-2.3eV)=29.1*10-31kg(6.78*10-19J)=1.22*106m/s

Thus, the stopping potential is 4.24 eV. The kinetic energy of the most energetic electrons ejected is 4.24 eV. The speed of theses electrons is 1.22*106m/s.

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