Modern Physics
Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 6, Problem 38P
To determine

The expression for average particle position; evaluate the results for mean position; the amplitude of oscillation for an electron in a well; the time to shuttle back and forth in a well; the same time classically for an electron.

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Answer to Problem 38P

Average position of particle is x0+Acos(Ωt) ; the mean position value is 0.5nm ; the amplitude value is 0.18nm ; time required is equal to 3.68×1015s ; classical time required is equal to 3.47×1015s .

Explanation of Solution

Write the expression for average position.

  x=x|Ψ|2dx

Substitute C{ψ1eiE1t/+ψ2eiE2t/} for ψ and C{ψ1*e+iE1t/+ψ2*e+iE2t/} for ψ* in above expression.

  x=C2x{ψ1*e+iE1t/+ψ2*e+iE2t/}{ψ1eiE1t/+ψ2eiE2t/}dx        (I)

Write the expression for mean position.

  x0=12(x|ψ1|2dx+x|ψ2|2dx)        (II)

Write the expression for amplitude of oscillation.

  A=xψ1*ψ2dx        (III)

Write the expression for energy of particle.

  En=n2h22mL2        (IV)

Write the expression for time required.

  T=hE2E1        (V)

Write the expression for velocity in terms of kinetic energy.

  v=(2Em)1/2        (VI)

Write the expression for classical time required.

  t=2Lv        (VII)

Conclusion:

For last two integrals, the integral value of both is same as the functions ψ1(x) and ψ2(x) is real.

Substitute 1/2 for C and simplify the equation (I).

  x=C2{x|ψ1|2dx+x|ψ2|2dx+eiΩtxψ1*ψ2dx+e+iΩtxψ2*ψ1dx}=12{(x|ψ1|2dx+x|ψ2|2dx)+eiΩtA+e+iΩtA}=12(x|ψ1|2dx+x|ψ2|2dx)+A2(eiΩt+e+iΩt)=x0+Acos(Ωt)

Thus, average position of particle is x0+Acos(Ωt) .

The average position of particle for any stationary state of well is L/2 .

Substitute L/2 for x|ψ2|2dx in equation (II).

  x0=12(L2+L2)=L2=0.5nm

Thus, the mean position value is 0.5nm .

  A=2L0Lxsin(πxL)sin(2πxL)dx=1L[L3π{3sin(πxL)sin(3πxL)}]0L1L(L3π)0L{3sin(πxL)sin(3πxL)}=0+1L(L3π)2[{9cos(πxL)cos(3πxL)}]0L=0.18nm

Thus, the amplitude value is 0.18nm .

Substitute 1.24 keVnm/c for h, 511 keV/c2 for m and 1nm for L in equation (IV).

      E1=(1.24 keVnm/c)28(511 keV/c2)(1 nm)2=0.376 eV

Substitute 1.24 keVnm/c for h, 511 keV/c2 for m, 2 for n and 1nm for L in equation (IV).

      E1=(2)2(1.24 keVnm/c)28(511 keV/c2)(1 nm)2=1.51 eV

Substitute 1.51 eV for E2 , 0.376 eV for E1 and 4.136×1015 eVs for h in equation (V).

  T=4.136×1015 eVs1.124 eV=3.68×1015s

Thus, time required is equal to 3.68×1015s .

Calculate the kinetic energy of electron.

  E=E1+E22=0.94eV

Substitute 511 keV/c2 for m and 0.94 eV  for E in equation (VI).

  v=(2(0.94eV)511 keV/c2)1/2=(1.92×103)c

Substitute 1 nm for L and  (1.92×103)c  for v  in equation (VII).

  t=2(1nm)(1.92×103)c=3.47×1015s

Thus, classical time required is equal to 3.47×1015s .

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