Respuesta :
Answer;
= 2.346 x 10^12 electrons.
Explanation and solution;
The energy of each photon can be obtained using Planck's equation:
E = hν = (hc)/λ
= (6.626 x 10^-34 J s)(2.9979 x 10^8 m/s) / (233 x 10^-9 m)
= 8.525 x 10^-19 Joules
The maximum kinetic energy for each freed electron will be;
= 8.525 x 10-19 J - 6.94 x 10^-19 J
= 1.585 x 10^-19 J
The number of 233 nm photons in the 2.00 x 10-6 J burst of light will be;
= 2.00 x 10^-6 J / 8.525 x 10^-19 J
= 2.346 x 10^12 photons.
Each photon can free at most one electron, so the maximum number of freed electrons will be 2.346 x 10^12 electrons.
The maximum possible kinetic energy of the emitted electrons is 2.379 x 10⁻¹⁹ J.
The maximum number of electron that can be ejected is 1.248 x 10¹³ electrons.
The given parameters;
- wavelength of the light, λ = 233 nm
- work function of titanium, φ = 3.84 ev
The maximum possible kinetic energy of the emitted electrons is calculated by applying Einstein photoelectric equation as follows;
[tex]E = K.E_{max} + \ \Phi[/tex]
where;
- E is the energy of the incident light
E = hf
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{233\times 10^{-9}} \\\\E = 8.531 \times 10^{-19} \ J[/tex]
The maximum possible kinetic energy of the emitted electrons;
[tex]K.E_{max}= E \ - \ \Phi\\\\K.E_{max}= (8.531\times 10^{-19}) \ - \ (3.84\times 1.602 \times 10^{-19})\\\\K.E_{max}= 2.379 \times 10^{-19} \ J[/tex]
The maximum number of electron that can be ejected when a light of 2 μJ incident on a metal.
1.602 x 10⁻¹⁹ J ------------------------ 1 electron
2 x 10⁻⁶ J ------------------------------- ?
[tex]n = \frac{2\times 10^{-6} }{1.602\times 10^{-19}} \\\\n = 1.248 \times 10^{13} \ electrons[/tex]
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