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Question: Show that Eq.38-24 is indeed a solution of Eq.38-22 by substituting ψ(x) and its second derivative into Eq.38-22 and noting

that an identity results.

Short Answer

Expert verified

It is shown that Eq. 38-24is indeed a solution of Eq. 38-22by substituting ψxand its second derivative into Eq.38-22.

Step by step solution

01

 Step 1: Identifying the data given in the question

Eq. 38-24is d2ψdx2+k2ψ=0

Eq.38-22isψx=Aeikx+Be-ikx

02

 Step 2: Concept used to solve the question.

The solution of the equation always satisfies that equation.

03

 Step 3: Finding the solution

The given function is

ψx=Aeikx+Be-ikx

Differentiating the function with respect to x

dψxdx=ddxAeikx+Beikxdψdx=ikAeikx-ikBeikx

Differentiating again with respect to x

d2ψdx2=ddxikAeikx-ikBeikxd2ψdx2=ik2Aeikx+-ik2Beikxd2ψdx2=-k2Aeikx-k2Beikx

Now substituting the value of d2ψdx2and ψxin the equation Eq.38-24

d2ψdx2+k2ψ=-k2Aeikx-k2Beikx+k2Aeikx+Beikxd2ψdx2+k2ψ=-k2Aeikx-k2Beikx+k2Aeikx+k2Beikxd2ψdx2+k2ψ=0

Sinced2ψdx2and ψx satisfying the equation d2ψdx2+k2ψ

Therefore, they are the solution of the equation.

Hence, it is shown that Eq. 38-24is indeed a solution of Eq.38-22 by substituting ψx and its second derivative into Eq.38-22 .

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Most popular questions from this chapter

Consider a collision between an x-ray photon of initial energy 50.0keVand an electron at rest, in which the photon is scattered backward and the electron is knocked forward.

(a) What is the energy of the backscattered photon?

(b) What is the kinetic energy of the electron?

Question: The function φ(x)displayed in Eq. 38-27 can describe a free particle, for which the potential energy is U(x)=0 in Schrodinger’s equation (Eq. 38-19). Assume now that U(x)=U0=a constant in that equation. Show that Eq. 38-27 is a solution of Schrodinger’s equation, with k=2πh2m(E-U0)giving the angular wave number k of the particle.

Question:For the arrangement of Figs. 38-14 and 38-15, electrons in the incident beam in region 1 have a speed of 1.60×107m/sand region 2 has an electric potential of V2-500V. What is the angular wave number in (a) region 1 and (b) region 2? (c) What is the reflection coefficient? (d) If the incident beam sends 3.00×109electrons against the potential step, approximately how many will be reflected?

Fig 38-14


Fig 38-15

Consider a balloon filled with helium gas at room temperature and atmospheric pressure. Calculate (a) the average de Broglie wavelength of the helium atoms and (b) the average distance between atoms under these conditions. The average kinetic energy of an atom is equal to 3/2kT, wherek is the Boltzmann constant. (c) Can the atoms be treated as particles under these conditions? Explain.

X rays of wavelength 0.0100 nm are directed in the positive direction of an x axis onto a target containing loosely bound electrons. For Compton scattering from one of those electrons, at an angle of , what are

(a) the Compton shift,

(b) the corresponding change in photon energy,

(c) the kinetic energy of the recoiling electron, and

(d) the angle between the positive direction of the x axis and the electron’s direction of motion?

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