# Sol-Ch41 - CHAPTER 41 E xercises 1 = h(2mk 1 2 = 4.909 10 1 9/K 1 2 w here K is in joules 1 e V = 1.602 10 1 9 J find = 1.23 nm/K 1 2 f or K in eV

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1. λ = h/(2mk) 1/2 = 4.909 ×  10 –19 /K 1/2 where K is in joules. 1 eV = 1.602 ×  10 –19 J; find λ = 1.23 nm/K 1/2 for K in eV. 2. λ = h/(2meV) 1/2 = 1.226 ×  10 –9 /V 1/2 m = (1.5/V) 1/2 nm 3. λ = (1.5/120) 1/2 = 0.112 nm 4. From Exercise 1, λ e = 1.23/(K) 1/2 = 0.87 nm λ p = hc/E = 622 nm 5. λ = h/mv: (a) 0.397 nm; (b) 0.397 pm 6. λ = h/(2mK) 1/2 = 0.143 nm 7. λ = h/mv = 6 ×  10 –32 m, a = λ /sin5 ° = 7.6 ×  10 –30 m, no. 8. Photon: E = hc/ λ = 248 eV; Electron: p = h/ λ and K = p 2 /2m = 0.061 eV 9. E = 1240 eV/ λ (nm): (a) 12.4 keV; (b) 1.24 geV 10. v = h/m λ = 1.2 km/s 11. λ = h/(2meV) 1/2 , so V = h 2 /(2me) λ 2 = 82 kV 12. Dsin Ф = λ = (1.5/V) 1/2 , thus 0.215sin Ф = 0.152, so Ф = 45 ° . 13. λ = 1.23/(K) 1/2 : (a) λ = 1.23/(80) 1/2 = 0.138 nm; (b) λ = 1.23/(100) 1/2 = 0.123 nm 14. v = h/m λ = 7.27 ×  10 6 m/s 15. (a) v = h/m λ = 1.37 ×  10 7 m/s; (b) v = (ke 2 /mr 1 ) 1/2 = 2.18 ×  10 6 m/s 16. 1/2 mv 2 = ke 2

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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-Ch41 - CHAPTER 41 E xercises 1 = h(2mk 1 2 = 4.909 10 1 9/K 1 2 w here K is in joules 1 e V = 1.602 10 1 9 J find = 1.23 nm/K 1 2 f or K in eV

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