Q.Consider a $20\ \text{W}$ bulb emitting light of wavelength $5000\ \text{\AA}$ and shining on a metal surface kept at a distance $2\ \text{m}$. Assume that the metal surface has work function of $2\ \text{eV}$ and that each atom on the metal surface can be treated as a circular disk of radius $1.5\ \text{\AA}$.

Concept: Photon energy and the quantum nature of light.

1. Photons per second. Photon energy $E=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{5000\times10^{-10}}=3.98\times10^{-19}\ \text{J}$. Rate $n=\dfrac{P}{E}=\dfrac{20}{3.98\times10^{-19}}\approx 5.0\times10^{19}\ \text{s}^{-1}.$
2. Emission? $E=3.98\times10^{-19}\ \text{J}=2.49\ \text{eV}>\phi=2\ \text{eV}$, so yes, photoelectric emission occurs.
3. Classical fill time. Intensity at 2 m: $I=\dfrac{P}{4\pi r^2}=\dfrac{20}{4\pi(2)^2}=0.398\ \text{W/m}^2$. Disk area $A=\pi(1.5\times10^{-10})^2=7.07\times10^{-20}\ \text{m}^2$. Power on one disk $=IA=2.81\times10^{-20}\ \text{W}$. Time to gather $\phi=3.2\times10^{-19}\ \text{J}$: $t=\dfrac{3.2\times10^{-19}}{2.81\times10^{-20}}\approx 11.4\ \text{s}.$
4. Photons in that time. Fraction intercepted $=\dfrac{A}{4\pi r^2}=1.41\times10^{-21}$; rate on a disk $=n\times$fraction$=0.071\ \text{s}^{-1}$; in $11.4$ s that is $\approx 0.8\ (\sim 1)$ photon. …

The idea

This problem contrasts the classical wave picture (energy spread continuously over a wavefront, so a tiny atom must slowly "collect" it) with Einstein's photon picture (energy arrives in whole quanta $hc/\lambda$). The classical estimate gives an absurd multi-second delay, which experiment flatly contradicts — that is the whole point.

(i) Photons emitted per second

Photon energy:

With all 20 W converted to light,

(ii) Will there be photoelectric emission?

Convert the photon energy to eV:

Since $E = 2.49\ \text{eV} > \phi = 2\ \text{eV}$, a single photon carries more than the work function, so yes — photoelectric emission takes place.

(iii) Classical time for a disk to collect $\phi$

Treat the bulb as an isotropic source. Intensity at $r = 2$ m:

Area of one atomic disk ($a = 1.5\ \text{Å}$):

Power falling on one disk:

Energy needed $\phi = 2\ \text{eV} = 3.2\times10^{-19}\ \text{J}$, so

(iv) Photons received by the disk in that time

Fraction of the emitted photons landing on one disk:

Photons per second on the disk: …

Method: Testing the Classical Wave Picture Against the Photon Picture (Instantaneous Emission Problems)

This class of question ("how could the photoelectric effect happen instantaneously") is answered by deliberately computing what classical wave theory would predict, and showing it disagrees badly with what a quantized-photon picture predicts and with experiment — the contradiction itself is the answer.

Steps

Step 1: Compute the photon emission rate from the source's power

This tells you how many discrete energy packets the source is putting out per second.

Step 2: Check whether a single photon already exceeds the work function

Convert $E_\gamma$ to eV and compare directly to $\phi$. If $E_\gamma>\phi$, the photon picture predicts immediate emission the instant one photon is absorbed — no waiting required.

Step 3: Compute the classical "accumulation time" for comparison

Classically, treat the source's energy as spread continuously over the sphere it illuminates. Find the intensity at the target's distance ($I=P/4\pi r^2$), the power falling on one atomic-sized area ($P_{\text{atom}}=I\times A_{\text{atom}}$), and the time needed to accumulate $\phi$ worth of energy at that (tiny) continuous rate:

$t = \frac{\phi}{P_{\text{atom}}}$ …