{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

KaSap Electronic Materials_Chapter_3_problem_solutions

# KaSap Electronic Materials_Chapter_3_problem_solutions -...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 3 Solutions 3.1 a. Given: power P = 1000 W and frequency υ = 700 × 10 3 s -1 . Then the photon energy is E ph = h υ = (6.626 × 10 -34 J s)(700 × 10 3 s -1 ) E ph = 4.638 × 10 -28 J/photon or 2.895 × 10 -9 eV/photon The number of photons emitted from the antenna per unit time ( N ph ) is therefore: J 10 638 . 4 W 1000 28 - × = = ph E P ph N = 2.16 × 10 30 photons per second b. Average intensity I average = 1000 W/m 2 and maximum wavelength λ max = 800 nm. The photon energy at λ max is, E ph = hc / λ max = (6.626 × 10 -34 J s)(3.0 × 10 8 m s -1 )/(800 × 10 -9 m) E ph = 2.485 × 10 -19 J or 1.551 eV The photon flux Γ ph is the number of photons arriving per unit time per unit area, J 10 485 . 2 m W 1000 19 2 average × = = Γ ph E I ph = 4.02 × 10 21 photons m -2 s -1 For the electric field we use classical physics. I average = (1/2) c ε o E 2 , so that ( ) ( )( ) 1 12 1 8 2 average m F 10 854 . 8 s m 10 0 . 3 m W 1000 2 2 × × = = o c ε I E = 868 V m -1 c. If each photon gives rise to one electron then the current density J is: J = e Γ ph = (1.602 × 10 -19 C)(4.02 × 10 21 m -2 s -1 ) = 644 A m -2 3.5 Photon energy, E ph = hc / λ = (6.63 × 10 -34 J s)(3.0 × 10 8 m/s)/(0.02 × 10 -9 m) E ph = 9.945 × 10 -15 J or 62.15 keV Photons per unit, Φ = P λ /hc = (0.1×10 -6 J cm -2 ) (0.02×10 -9 m)/(6.63×10 -34 J s)(3×10 8 ms -1 ) Φ = 1 × 10 7 photons cm -2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3.7 Solution a. We are given λ max = 420 nm. The work function is then: Φ = h υ o = hc / λ max = (6.626 × 10 -34 J s)(3.0 × 10 8 m s -1 ) / (420 × 10 -9 m) Φ = 4.733 × 10 -19 J or 2.96 eV b. Given λ = 300 nm, the photon energy is then: E ph = h υ = hc / λ = (6.626 × 10 -34 J s)(3.0 × 10 8 m s -1 )/(300 × 10 -9 m) E ph = 6.626 × 10 -19 J = 4.14 eV The kinetic energy KE of the emitted electron can then be found: KE = Φ - E ph = 4.14 eV - 2.96 eV = 1.18 eV c. The photon flux Γ ph is the number of photons arriving per unit time per unit area. If I light is the light
This is the end of the preview. Sign up to access the rest of the document.