20111ee2_1_2011_EE2_HW1 Solution

20111ee2_1_2011_EE2_HW1 Solution - Solutions to HW #1 EE2...

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Solutions to HW #1 EE2 Winter 2011 Problem 1 The de Broglie wavelength of a particle (mass m, velocity v) is given by mv h p h = = λ , where h is the Planck’s constant ( = 6.626x10 -34 Js) Therefore, λ = 6.626x10 -34 Js/{1.5x10 -3 kg x (10 x 5/18)ms -1 } = 1.590x10 -31 m (incredibly small!!) Problem 2 E = 1x10 -18 J We also know that E = p 2 /2m and the mass of an electron, m e = 9.11x10 -31 kg Therefore, p = (2 x 9.11x10 -31 x1x10 -18 ) = 1.350x10 -24 kgms -1 As in problem 1, λ = 6.626x10 -34 Js/1.350x10 -24 kgms -1 = 4.909x10 -10 m 4.9 A° Problem 3 From the solution of a particle in a infinite potential box , we know that the quantized values of the kinetic energy and the various components of momentum are: 2 2 2 2 222 8 ,, y x z xyz x xy yz n n n h E mL L L hh pn p LL ⎛⎞ =+ + ⎜⎟ ⎝⎠ === z z h n L where n x , n y , n z are integers (quantum numbers) and L x , L y , L z are the lengths of the sides of the box Therefore, for a state described by (2, 3, 2), we have p x = h/L x , p y = 3h/2L y , p z = h/L z and E = h 2 /8m x (4/L x 2 + 9/L y 2 + 4/L z 2 ) which are evaluated as below: p x = 6.626x10 -34 Js/1x10 -9 m = 6.626x10 -25 kgms -1 p y = 3 x 6.626x10 -34 Js/2x2x10 -9
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20111ee2_1_2011_EE2_HW1 Solution - Solutions to HW #1 EE2...

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