ECE331_Wi06_HW3_sol

ECE331_Wi06_HW3_sol - ECE331 HW#3 Solution 1 Streetman...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE331 HW#3 Solution 1. Streetman, 2.5 (a)  h 6.63 × 10 −34 kg * m ∴ ∆p x = = = = 1.06 × 10 − 24 −10 ∆x 2π∆x 2π 1 × 10 s (b)  h 4.14 × 10 −15 1 ∴ ∆t = = = = 6.59 × 10 −16 ∆E 2π∆E 2π (1) s ( ) 2) Streetman, 2.6 1 E = mv 2 −19 2 , 1eV = 1.6 × 10 λ= h = mv h m 2E m = h 2 Em = 6.63 × 10 −34 2 ∗ 9.11 × 10 −31 E − 1 2 = 4.91 × 10 −19 E − 1 2 i) For 100eV 4.91 × 10 −19 4.91 × 10 −19 λ100 eV = = = 1.23 × 10 −10 m = 1.23Α −19 E 100 * 1.6 × 10 ii) For 12keV 4.91 × 10 −19 4.91 × 10 −19 λ12 keV = = = 1.12 × 10 −11 m = 0.112 Α 3 −19 E 12 × 10 * 1.6 × 10 λ = 0.5um = 5000 Α , an electron microscope has much higher resolution For visible light light 3) Streetman, 2.8 At x<0 nm, the probability of finding the electron is zero. At x>6 nm, the probability is non-zero. At x>6 nm, if the electron was described by classical mechanics, this probability is zero. 4) Streetman, 2.9 Ψ = Αe j (10 x +3 y − 4t ) ∞ *  ∂ ∫∞Ψ j ∂x Ψ∂x − = px = ∞ * ∫ Ψ Ψ∂x (i) −∞ ( )  ∞ * − j (10 x +3 y −4t ) ∂ Αe j (10 x +3 y − 4t ) ∫Α e ∂x j ∂x −∞ ∞ * − j (10 x + 3 y − 4 t ) ∫Α e Αe j (10 x +3 y −4t )∂x −∞ ∞  ( j10) ∫ Α * e − j (10 x +3 y −4t ) Αe j (10 x +3 y − 4t )∂x j = ∞ −∞ = (10 ) * − j (10 x + 3 y − 4 t ) j (10 x + 3 y − 4 t ) Αe ∂x ∫Α e −∞ 1 ∴ p x = (10 ) (ii) ∞ pz  ∂ ∫∞Ψ j ∂z Ψ∂x = − ∞ = * ∫ Ψ Ψ∂x * −∞ ∞  ∞ * − j (10 x +3 y −4t ) ∫Α e (0)∂x j −∞ * − j (10 x + 3 y − 4 t ) ∫Α e Αe j (10 x + 3 y − 4 t ) =0 ∂x −∞ ∴ p z = 0 , because Ψ has no z dependence. (iii) ( ) ∂  ∞ * − j (10 x + 3 y − 4t ) ∂ Αe j (10 x + 3 y − 4t ) ∂x ∫ Ψ − j ∂t Ψ∂x − j −∫∞Α e ∂t −∞ = E= ∞ ∞ * * − j (10 x + 3 y − 4 t ) Ψ Ψ∂x Αe j (10 x + 3 y − 4t )∂x ∫ ∫Α e ∞ * −∞ −∞ ∞  − (− 4 j ) ∫ Α*e − j (10 x + 3 y − 4t )Αe j (10 x + 3 y − 4t )∂x j −∞ = = (4 ) ∞ * − j (10 x + 3 y − 4 t ) j (10 x + 3 y − 4 t ) Αe ∂x ∫Α e −∞ ∴ E = (4 ) 2 3 ∴{4 p x + 2 p z + 7 E m} = 4 ⋅ 100 ⋅  2 + 7 ⋅ 4 ⋅  / m = 400 2 + 28 / m 5. Streetman, 2.11 ( ) 2 n 2π 2  2 6.63 ×10−34 En = = n 2 = 6.03 ×10−20 n 2 −31 −10 2mL 8 9.11×10 10 ×10 ( )( ) eV = 0.377eV 1.6 × 10 −19 J E 2 = 0.377n 2 = 0.377(4) = 1.508eV E1 = 6.03 × 10 − 20 J E 3 = 0.377 n 2 = 0.377(9 ) = 3.393eV 6. ′ Ψ2 ( x) = exp{− 2k 2 d } ′ k2 = 2m(Vo − E )  2 ( )( ) 121 mv = 9.11 × 10 −31 10 5 2 2 E= 2 = 4.555 × 10 − 21 Vo = 3E = 1.3665 × 10 −20 J ′ k2 = 2m(Vo − E )  = 2π 2(9.11 × 10 −31 )(9.11 × 10 − 21 ) kg = 1.22 × 10 9 −34 s 6.63 × 10 Thus, Ψ2 ( x) = exp{− 2(1.22 × 10 9 )d } For 10A (Put in KMS) Ψ2 ( x) = exp{− 2(1.22 × 10 9 )(10Α )} = exp{− 2(1.22 × 10 9 )(10 × 10 −10 )} = 0.0872 For 100A Ψ2 ( x) = exp{− 2(1.22 × 10 9 )(100Α )} = exp{− 2(1.22 × 10 9 )(100 × 10 −10 )} = 2.53 × 10 −11 7. Vo T = 1 + sinh 2 (ka ) 2 2 (Vo ) − (2 E − Vo ) 2m(Vo − E ) k=  = 2π 2(9.11 × 10 −31 )(2 − 0.25)(1.6 × 10 −19 4.14 × 10 −15 = 6.76 × 10 9 kg s Thus 4 T = 1 + 2 sinh 2 (2.71a ) 2 (2 ) − (2(0.25) − 2 ) =1 3.28 sinh 2 6.76 × 10 9 a [ ( −1 ( )] −10 −10 Graph for 2 × 10 ≤ a ≤ 20 × 10 Α ) 4 sinh 2 6.76 × 10 9 a = 1 + 1.75 (see graph below) 3 −1 Tunneling Probability 1 0.1 0.01 0.001 Probability 0.0001 1E-05 1E-06 1E-07 1E-08 1E-09 1E-10 1E-11 1E-12 2.00E-12 4.00E-12 6.00E-12 8.00E-12 1.00E-11 1.20E-11 Barrier Width (m ) 4 1.40E-11 1.60E-11 1.80E-11 2.00E-11 5 ...
View Full Document

This note was uploaded on 10/16/2011 for the course ECE 331 taught by Professor Rajan during the Spring '09 term at Ohio State.

Ask a homework question - tutors are online