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Chapter 5: Q33E (page 188)

Verify that Asin(kx) + Bcos(kx)is a solution of equation (5 - 12).

Short Answer

Expert verified

The wave function equation satisfies the Schrodinger Equation.

Step by step solution

01

Wave function equation and Schrodinger Equation.

Schrodinger Equation,

d2ψdx2= -k2ψ(x),k =2mEh2...................1

Wave function equation,

ψ== A sin(kx)+B cos(kx)........1

02

Substituting (2) in to (1).

Substitute the values

d2ψ(x)dx2= -k2A sin(kx) -k2B cos(kx)dψ(x)dx= k A cos(kx) - k B sin(kx)d2ψ(x)dx2= -k2A sin(kx) -k2=k2ψxSo, thatwave function equation satisfies the Schrodinger Equation.

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Most popular questions from this chapter

When is the temporal part of the wave function 0 ? Why is this important?

a) Taking the particle’s total energy to be 0, find the potential energy.

(b) On the same axes, sketch the wave function and the potential energy.

(c) To what region would the particle be restricted classically?

The figure shows a potential energy function.

(a) How much energy could a classical particle have and still be bound?

(b) Where would an unbound particle have its maximum kinetic energy?

(c) For what range of energies might a classical particle be bound in either of two different regions?

(d) Do you think that a quantum mechanical particle with energy in the range referred to in part?

(e) Would be bound in one region or the other? Explain.

Consider a particle of mass mand energy E in a region where the potential energy is constant U0. Greater than E and the region extends tox=+∞

(a) Guess a physically acceptable solution of the Schrodinger equation in this region and demonstrate that it is solution,

(b) The region noted in part extends from x = + 1 nm to +∞. To the left of x = 1nm. The particle’s wave function is Dcos (109m-1 x). Is also greater than Ehere?

(c) The particle’s mass m is 10-3 kg. By how much (in eV) doesthe potential energy prevailing from x=1 nm to U0. Exceed the particle’s energy?

Show that the uncertainty in the position of a ground state harmonic oscillator is 1/2ℏ2/mk1/4.
See all solutions

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