Question

In the photoelectric effect, photoelectrons begin leaving the surface at essentially the instant that light is introduced. If light behaved as a diffuse wave and an electron at the surface of a material could be assumed localized to roughly the area of an atom, it would take far longer. Estimate the time lag. assuming a work function of4eV, an atomic radius of approximately0.1nm,and a reasonable light intensity of$0.01W/m^2$

Short answer

The time lag of the photoelectron is $2038\text{s}$.

Step-by-step solution

Given data

Work function $=4\text{eV}$

Radius $=0.1\text{nm}$

Light intensity $=\text{0.01}\text{W}·{\text{m}}^{-2}$.

Formula used  

Power,$P=\frac{E}{t}$

Intensity, $I=\frac{P}{A}$

Calculate byIntensity and Power

Intensity is given by,

$I=\frac{P}{A}$

Where, P is the power and A is the area.

Power can be written in the form,

$P=\frac{E}{t}$

Where, t is the time and E is the energy.

Combining the above equations and substituting $E=\varphi$ (work function) will give,

$$\begin{gathered} I=\frac{E}{tA} \\ I=\frac{\varphi }{tA} \end{gathered}$$

Calculate the time

Therefore,

$\begin{array}{rcl}t& =& \frac{\varphi }{AI}\\ & =& \frac{4\times 1.6\times {10}^{-19}J}{\pi \times {\left(0.1\times {10}^{-9}m\right)}^2\times 0.01W/m^{-2}}\\ & =& 2038\text{s}\end{array}$

Conclusion 

The time lag of the photoelectron is $2038\text{s}$.