Consider a `20W` bulb emitting light of wavelength `5000Å` and shinning on a metal surface kept at a distance `2m`. Assume that the metal surface has work function of `2eV` and that each atom on the metal surface can be treated as a circular disk of radius `1.5Å`.
(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses] (ii) Will there be photoelectric emission? (iii) How much time would be required by the atomic disk to receive energy equal to work function `2eV`? (iv) How many photons would atomic disk receive within time duration calculated in (iii) above? (v) Can you explain how photoelectric effect was observed instantaneously? [Hint : Time calculated in part (iii) is from classical consideration and you may further take the target of surface area say `1cm^(2)` and estimate what would happen?]

Explanation

Given, `P=200W,lamda=5000Å=5000xx10^(-10)m`
`d=2m,phi_(0)=2eV,r=1.5A=1.5xx10^(-10)m`
(i) Number of photon emitted by bulb per second is n' `=(p)/(hc//lamda)=(plamda)/(hc)`
`" "=(20xx(5000xx10^(-10)))/((6.62xx10^(-34))xx(3xx10^(8)))`
`implies" "=5xx10^(19)s^(-1)`
(ii) Energy of the incident photon
`(hc)/(lamda)=((6.62xx10^-34)(3xx10^(8)))/(5000xx10^(-10)xx1.6xx10^(-19))`
`" "=2.48` eV
As this energy is greater than e eV (i.e., work function of metal surface), hence photolectric emission takes place.

(ii) Let `Deltat` be the time spent in getting the energy `phi`= (work functiojn of metal). Consider the figure,
`" "(P)/(4pid^(2))xxpir^(2)Deltat=phi_(0)`
`implies" "Deltat=(4phi_(0)d^(2))/(Pr^(2))`
`" "=(4xx(2xx1.6xx10^(-19))xx2^(2))/(20xx(1.5xx10^(-10))^(2))~~28.4s`
(iv) Number of photons receivbed by atomic disc in time `Deltat` is
`" "N=(n'xxpir^(2))/(4pid^(2))xxDeltat`
`implies" "=(n'r^(2)Deltat)/(4d^(2))`
`" "=((5xx10^(19))xx(1.5xx10^(10))^(2)xx28.4)/(4xx(2)^(2))~~2`
(v) As time of emiccion of electrons is 11.04 s.
Hence, the photoelectric emission is not instantaneous in this problem.
In photoelectric emission, there is an collision between incident photon and free electron of the metal surface, which lasts for ever very short interval of time `(~~10^(-9)S),` hence we say photoelectric emission is instantaneous.