{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20111ee2_1_2011_EE2_HW1 Solution

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

This preview shows pages 1–2. Sign up to view the full content.

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 2 2 2 8 , , 2 2 2 y x z x y z x x y y z x y n n n h E m L L L h h p n p n p L L = + + = = = 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

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

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

{[ snackBarMessage ]}