Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
Question
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Chapter 41, Problem 59CP

(a)

To determine

Prove that the reflection coefficient for the given condition is R=(k1k2)2(k1+k2)2.

(a)

Expert Solution
Check Mark

Answer to Problem 59CP

Proof for the reflection coefficient for the given condition is R=(k1k2)2(k1+k2)2 shown below.

Explanation of Solution

Write the Schrodinger’s equation.

    2ψx2=2m2(EU)ψ                                                   (I)

Here, ψ is the wavefunction, m is the mass, E is the energy, U is the potential energy and is h/2π, h is Plank’s constant

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 41, Problem 59CP

The solutions of the equation I for the region I in the above figure is

    ψ1=Aeik1x+Beik1x                                                 (II)

The solutions of the equation I for the region II in the above figure is

    ψ2=Ceik2x                                                 (III)

Here, A, B and C are the constants.

Check the solution for the region I satisfies the equation I.

    2ψ1x2=2m2Eψ12x2(Aeik1x)+2x2(Beik1x)=2m2E(Aeik1x+Beik1x)k12(Aeik1x)k12(Beik1x)=2m2E(Aeik1x+Beik1x)k12(Aeik1x+Beik1x)=2m2E(Aeik1x+Beik1x)

The above equation will be true only if the k12=2m2E, then substitute the above value in the following energy equation.

    E=p22m=(k1)22m

Here, p is the momentum.

Rewrite the above equation to find the value of k1.

    k1=2mE                                                 (IV)

Therefore the equation I is satisfied for the region I.

Check the equation I for the region II.

    2ψ2x2=2m2(EU)ψ22x2(Ceik2x)=2m2(EU)(Ceik2x)k22(Ceik2x)=2m2(EU)(Ceik2x)

The above equation will be true only if the k22=2m2(EU), then substitute the above value in the following energy equation.

    E=p22m+U=(k2)22m

Rewrite the above equation to find the value of k2.

    k2=2m(EU)                                                 (V)

Therefore the equation I is satisfied for the region II.

Then apply the boundary conditions such as matching the function and derivatives at x=0.

    (ψ1)0=(ψ2)0  gives A+B=C

    (dψ1dx)0=(dψ2dx)0 gives k1(AB)=k2C

From the above equations,

    B=1k2/k11+k2/k1A and C=21+k2/k1A                                                (VI)

Write the equation for probability.

    R=B2A2                                                 (VII)

Conclusion:

Substitute equation VI in VII.

    R=(1k2/k1)2(1+k2/k1)2=(k1k2)2(k1+k2)2

Therefore, the proof for the reflection coefficient for the given condition is R=(k1k2)2(k1+k2)2 shown above.

(b)

To determine

Find the probability of particle being reflected.

(b)

Expert Solution
Check Mark

Answer to Problem 59CP

The probability of particle being reflected is 0.0920.

Explanation of Solution

Write the equation for ration of k2 to k1.

    k2k1=EUE                                                  (VIII)

Conclusion:

Substitute 7.00 eV for E and 5.00 eV for U in equation VIII.

    k2k1=2.00 eV7.00 eV=0.535

Substitute 0.535 for k2k1 in the following probability equation.

  R=(1k2/k1)2(1+k2/k1)2=(10.535)2(1+0.535)2=0.0920

Therefore, the probability of particle being reflected is 0.0920.

(c)

To determine

Find the probability of particle being transmitted.

(c)

Expert Solution
Check Mark

Answer to Problem 59CP

The probability of particle being transmitted is 0.908.

Explanation of Solution

Write the equation for probability of transmitted.

    T=1R                                                  (IX)

Conclusion:

Substitute 0.0920 for R in equation IX.

    T=10.0920=0.908

Therefore, the probability of particle being transmitted is 0.908.

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Chapter 41 Solutions

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