Total energy emitted by the bulb in 1 second = 20J
Wavelength of incident radiation =
(i) Let number of photons emitted per second be ‘n’
(ii) Since, this value is greater than the work function of the metal, photoelectric mission will take place.
(iii) Let t be the time spent in giving energy equal to its work function (φ).
The total energy being supplied = 20W.
This energy is being spread in all directions, so the energy to one atom will be given by (‘d’ is the distance between the bulb and metal surface)
In time ‘t’, the energy equivalent to work function of metal is received, so
Putting d = 2m, r = 1.5 x 10-10 m and φ = 2eV, we get t = 11.3 s.
(iv) Let the number of photons received by atomic disc in time ‘t’ be N
(v) The time taken by an atom to receive energy equal to the work function of the metal would be same as time before emission of photoelectrons because emission will start only after breaching the work function energy level.
As time of emission of electrons is 11.3 s, the photoelectric effect is not instantaneous in this problem.
In photoelectric emission, there is a collision between the incident photon and free electrons of the metal surface which lasts for very short time (order of 10-9 s) and if the energy of one photon is greater than the work function, only then photoelectric emission takes place, thus, we say photoelectric emission is instantaneous.
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