1101_PartUniversity Physics Solution

# 1101_PartUniversity Physics Solution - Quantum Mechanics...

This preview shows pages 1–3. Sign up to view the full content.

Quantum Mechanics 40-11 sinh 2 LL ee L κκ κ = For 1, sinh 2 L e W and 1 22 00 0 16 ( ) 1 16 ( ) 16 ( ) L L Ue EU E T E E U e ⎡⎤ →+ = ⎢⎥ −− + ⎣⎦ For 00 0 1, 16 ( ) L E Ue −+ W 2 0 0 16 ( ) 16 1 , L L E E E Te U U ⎛⎞ →= ⎜⎟ ⎝⎠ which is Eq.(40.21). (b) 0 2( ) . Lm U E L = = So 1 L W when L is large (barrier is wide) or 0 UE is large. ( E is small compared to 0 . U ) (c) 0 ) ; mU E = = becomes small as E approaches 0 . U For small, sinh and 11 222 2 2 0 2 ) 4( ) ) UL Um L T E E =+ = (using the definition of ) Thus 1 0 2 2 1 4 ULm T E = 0 so 2 0 U E E and 1 2 2 2 1 4 EL m T = But 2 2 2 , mE k = = so 1 2 1, 2 kL T as was to be shown. EVALUATE: When L is large Eq.(40.20) applies and T is small. When 0 , T does not approach unity. 40.48. (a) 2 1 (( 1 2 ) ) (( 1 2 ) ) , 2 Em vn ω nh f == + = + = and solving for n , 2 2 30 34 1 1 (1/2)(0.020 kg)(0.360 m/s) 1 2 1.3 10 . 2 (6.63 10 J s)(1.50 Hz) 2 mv n hf =− = = × ×⋅ (b) The difference between energies is 34 34 (6.63 10 J s)(1.50 Hz) 9.95 10 J. ω hf × = × = This energy is too small to be detected with current technology 40.49. IDENTIFY and SET UP: Calculate the angular frequency ω of the pendulum and apply Eq.(40.26) for the energy levels. EXECUTE: 1 4 s 0.500 s T ππ ωπ = The ground-state energy is 34 1 34 0 (1.055 10 J s)(4 s ) 6.63 10 J. E × = × = 34 19 15 0 6.63 10 J(1 eV/1.602 10 J) 4.14 10 eV E × 1 2 n En = 1 1 1 2 n + + = The energy difference between the adjacent energy levels is 10 2 nn EE E E + Δ= − = = = = 33 15 1.33 10 J 8.30 10 eV ×= × EVALUATE: These energies are much too small to detect. Quantum effects are not important for ordinary size objects. 40.50. IDENTIFY: We model the electrons in the lattice as a particle in a box. The energy of the photon is equal to the energy difference between the two energy states in the box. SET UP: The energy of an electron in the n th level is 2 . 8 n E mL = We do not know the initial or final levels, but we do know they differ by 1. The energy of the photon, / , hc λ is equal to the energy difference between the two states. EXECUTE: The energy difference between the levels is 34 8 7 (6.63 10 J s)(3.00 10 m/s) 1.649 10 m hc E ×⋅ × = = × 18 1.206 10 J. × Using the formula for the energy levels in a box, this energy difference is equal to (1 ) ( 21 ) . 88 hh En n n mL mL − − =

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

View Full Document
40-12 Chapter 40 Solving for n gives 21 8 3 1 9 2 23 4 2 8 1 (1.206 10 J)8(9.11 10 kg)(0.500 10 m) 11 3 . 2( 6 . 6 2 6 1 0 J s ) EmL n h −− ⎛⎞ Δ× × × =+ = + = ⎜⎟ ×⋅ ⎝⎠ The transition is from 3 n = to 2. n = EVALUATE: We know the transition is not from the 4 n = to the 3 n = state because we let n be the higher state and 1 n the lower state. 40.51. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. SET UP: The given wave function is 22 / 2 00 () x xA e α ψ = and the Schrödinger equation is 2 .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

### Page1 / 5

1101_PartUniversity Physics Solution - Quantum Mechanics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online