Give Einstein's explanation of photoelectric effect.

Explanation

Einstein assumed that one photoelectron is liberated from a metal surface when one photon of suitable frequency is incident on it. Consider a photon of frequency v and energy hv incident on a photo sensitive metal. The energy of photon is spent in liberating electrons.
(i) Since the collision between an electron and a photon is treated as the collision between two particles, the process is instantaneous.
(ii) If `E = hv` represents the incident photon energy and .w. is the minimum energy required to just liberate the electron from the surface and K.E. is the kinetic energy of electron then
`E= w + K.E. `
or `hv = hv_(0) + "K.E." rarr (1)` This is known as the P.E. equation.
For K.E.`= 0, hv = hv_(0) = w.` Therefore for every metal surface there exists minimum energy of the electron above which electron is liberated. The corresponding frequency is called the threshold frequency and the minimum energy corresponding to v is known as the work function.
(iii) K.E. `=1/2mv^(2)`
`1/2mv^(2) = hv - hv_(0)`
`11/2 mv^(2) = h(v-v_(0))`

`therefore` K.E. and hence velocity of photoelectron depends on the frequency of incident radiation above the threshold frequency `(v_(0))` but not on the intensity of incident radiation.
(iv) For a particular negative potential of the anode, photoelectrons do not reach the anode. Hence P.E. current `= 0`. This -ve potential of the anode for which PEI` = 0` is known as stopping potential.

(v) For `vlt v_(0) , 1/2 mv^(2) = -ve `
which is absurd
`therefore` P.E.E. cannot take place for a frequency less than the threshold frequency.
(vi) Photo electrons will be ejected in large numbers when intense radiation is incident because intense radiation contains more photons resulting in more interactions.