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As an electron escapes energy equivalent to the work function ‘${\phi}$’ of the emitter substance is given up. So the photon energy ‘h f’ must be greater than or equal to ${\phi}$. If the ‘h f’ is greater than ${\phi}$ then the electron acquires some kinetic energy after leaving the surface. The maximum kinetic energy of the ejected photoelectron is given by;
K.Emax = ½ m v2max = h f – F ……………… (i), where mv2max = maximum velocity and mass. This is the Einstein’s photoelectric equation.
If the photon energy is just equivalent to work function then, mv2max = 0, at this juncture the electron will not be able to move hence no photoelectric current, giving rise to a condition known as cut-off frequency, hfco = ${\phi}$………………. (ii)
Also the p.d required to stop the fastest photoelectron is the cut-off potential, V cowhich is given by E = e Vco electron volts, but this energy is the maximum kinetic energy of the photoelectrons and therefore,
½ m v2max = e Vco ………….. (iii).
Combining equations (i), (ii) and (iii), we can write Einstein’s photoelectric equation as,
eVco = hf – hfco ………………….. (iv)
NOTE: -- Equations (i) and (iv) are quite useful in solving problems involving photoelectric effect.
Mutiso answered the question on October 17, 2018 at 04:56
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