Sol-Ch40 - CHAPTER 40 E xercises 1. Use Eq. 40.1. (a) 0.97...

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Exercises 1. Use Eq. 40.1. (a) 0.97 mm; (b) 0.97 μm; (c) 0.29 nm 2. Use Eq. 40.1. (a) 6170 K; (b) 8280 K 3. From Eq. 40.1, T = 4140 K to 7250 K 4. R = σ˚(T 4 – T o 4 ). Convert ˚C to K by adding 273. (a) 1.5 ×  10 6 W/m 2 ; (b) 140 W/m 2 ; (c) 430 w/m 2 5. Power = AR = σT 4 (4π r2 ) = 3.8 ×  10 26 W 6. Power = RA = σ(T 4 – T o 4 )(2πrL) = 2280 W 7. From Eq. 40.1. λ = 9.66 μm 8. E = hf = 0.21 eV 9. ∆E = P∆t – nhf, thus n = 6.03 ×  10 29 each second. 10. (a) E = hc/ λ = (4.14 ×  10 –5 eV.s)(3 ×  10 8 m/s) λ = 1240 ×  10 –9 eV/ λ where λ is in meters. Thus, E = 1240/ λ is in nm. (b) E = 1.77 to 3.10 eV 11. (a) K m = hf – Ф = 0.855 eV; (b) hc/ λ = 4.97 ×  10 –19 J, so N = 3 ×  10 –11 W/m 2 /(hc/ λ ) = 6.0 ×  10 7 m –2 s – 1 12. (a) P = I(πr 2 ) = 9.8 ×  10 – 18 W; (b) N = P/hf = 25 photons/s. 13. K m = hc(1/ λ – 1/ λ o ) = 1.33 ×  10 – 19 J; v = (2K/m) 1/2 = 5.4 ×  10 5 m/s 14. See Exercise 10, E = 1240 eV/ λ , for λ in nm. (a) 2.25 eV; (b) 4.14 ×  10 –7 eV; (c) 3.89 ×  10 –9 eV; CHAPTER 40
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(d) 1.75 ×  10 4 eV 15. (a) f = E/h = 2.66 ×  10 15 Hz; (b) From Exercise 10:E = 1240/ λ = 7.09 eV 16. From Exercise 10: λ = 1240/E = 443 nm, viole 17. hc/ λ = 3.61 ×  10 – 19 J. N = I/ht = 3.72 ×  10 21 m – 2 s – 1 18. E = hc/ λ = 3.14 ×  10 –19 J; N = P/E = 3.18 ×  10 15 photons/s. 19. (a) K m = hf – Ф = 0.8 eV; (b) eV o = hc/ λ Ф , so λ = 428 nm 20. (a) K m = hc/ λ Ф = 1.7 eV; (b) k m = eV o , so V o = 1.7V. 21. K m = hf – Ф , so Ф = 2.34 eV, then eV o = hc/ λ Ф leads to V o = 3.05 V 22. eV o = hf – hf o , thus f o = 1.33 ×  10 14 Hz 23. (a) E = hc/ λ = 3.32 ×  10 – 19 J, so N = (5 W)/E = 1.51 ×  10 19 s –1 ; (b) NA/4πr 2 = 20, where a = πd 2 /4. Find r = 652 km 24. K 1
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch40 - CHAPTER 40 E xercises 1. Use Eq. 40.1. (a) 0.97...

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