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?]

Options

A. The number of photons emitted by the bulb per second. [Assume no other losses] is `5xx10^(19)s^(-1)`
B. There will be photoelectric emission
C. Time required by the atomic disk to receive energy equal to work function `(2eV)` is `11.4s`
D. The number of photons would atomic disk receive within time duration calculated in `(iii)` above is `2`

Answer

A, B, C

Explanation

Given `P=20W,lambda=5000overset(0)(A)=5xx10^(-7) m`.
`d=2m,phi_(0)=2eV,r=1.5oversetA=1.5xx10^(-10)m`.
(A)Number of photons emitted by bulb per second is `n=P/((hc//lambda))=(Plambda)/(hc)=(20xx5xx10^(-7))/(6.62xx10^(-34)xx3xx10^(8))`
(B)Energy of incident photon, `E=(hc)/lambda`
=`((6.62xx10^(-34))(3xx10^(8)))/(5xx10^(-7)xx1.6xx10^(-19))=2.48eV`.
As `E gt phi_(0)(248eV gt 2 eV)`
hence photoelectric emission will take place.
(C)Energy emitted by the bulb in time `Delta t=P Delta t`
Energy falling on the disk i.e., `E=((PDeltat)/(4pid^(2)))(pir^(2))=((Pr^(2))/(4d^(2)))Delta t`
According to given condition,`E=phi_(0)`
`therefore ((Pr^(2))/(4d^(2)))Delta t=phi_(0),Deltat=(4phi_(0)d^(2))/(Pr^(2))`
`Deltat=((4)(2xx1.6xx10^(-19))(2)^(2))/((20)(1.5xx10^(-10))^2)=11.4s`
(D)Number of photons received by atomic disk in time `Deltat` is `N=n((pir^(2))/(4pid^(2)))xxDeltat=(nr^(2)Deltat)/(4d^(2))`
`=(5xx10^(19)xx(1.5xx10^(-10))xx11.4)/(4(2)^(2))=0.8~~1`