Chapter 3: Q16E (page 93)

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 $0.01 W / m^{2}$

Short Answer

Expert verified

The time lag of the photoelectron is $2038 s$ .

Step by step solution

Given data

Work function $= 4 eV$

Radius $= 0.1 nm$

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

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 $E = ϕ$ (work function) will give,

$I = \frac{E}{t A} I = \frac{ϕ}{t A}$

Calculate the time

Therefore,

$\begin{array}{rcl} t & = & \frac{ϕ}{A I} \\ = & \frac{4 \times 1.6 \times 10^{- 19} J}{π \times {(0.1 \times 10^{- 9} m)}^{2} \times 0.01 W / m^{- 2}} \\ = & 2038 s \end{array}$

Conclusion

The time lag of the photoelectron is $2038 s$ .

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Short Answer

Step by step solution

Given data

Formula used

Calculate byIntensity and Power

Calculate the time

Conclusion

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