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Modern Physics
Found in: Page 186
Modern Physics

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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Short Answer

Consider a particle bound in a infinite well, where the potential inside is not constant but a linearly varying function. Suppose the particle is in a fairly high energy state, so that its wave function stretches across the entire well; that is isn’t caught in the “low spot”. Decide how ,if at all, its wavelength should vary. Then sketch a plausible wave function.

The graph is plotted with energy and potential inside a potential well.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1 : Location of turning points

If potential rises higher than the particles total energy then the particle is stuck in the potential well and it oscillates between the turning points and it cannot escape. Hence the turning points in this condition is located thus the oscillator wave function stretches across the entire well with quantum states with node and antinode.

Step 2 : Relation with kinetic energy and wavelength

As the kinetic energy varies inversely with wavelength, amplitude becomes larger at the regions where kinetic energy is smaller and the wavelength will be shortened for region where kinetic energy is larger

For high potential inside a well , the particles wave function tunnels through the finite potential barrier and is finally brought to zero.

Step 3 : Graph

the plausible wavefunction is

Most popular questions for Physics Textbooks

What is the probability that a particle in the first excited (n=2) state of an infinite well would be found in the nuddle third of the well? How does this compare with the classical expectation? Why?

In the harmonic oscillators eave functions of figure there is variation in wavelength from the middle of the extremes of the classically allowed region, most noticeable in the higher-n functions. Why does it vary as it does?

In a study of heat transfer, we find that for a solid rod, there is a relationship between the second derivative of the temperature with respect to position along the rod and the first with respect to time. (A linear temperature change with position would imply as much heat flowing into a region as out. so the temperature there would not change with time).

∂2T(x,τ)∂x2=β∂T(x,τ)∂τδx

(a) Separate variables this assume a solution that is a product of a function of x and a function of t plug it in then divide by it, obtain two ordinary differential equations.

(b) consider a fairly simple, if somewhat unrealistic case suppose the temperature is 0 at x=0 and , and x=1 positive in between, write down the simplest function of x that (1) fits these conditions and (2) obey the differential equation involving x. Does your choice determine the value, including sign of some constant ?

(c) Obtain the full T(x,t) for this case.

Show thatΔp=0⇒p^ψ(x)=p¯ψ(x) that is, verify that unless the wave function is an Eigen function of the momentum operator, there will be a nonzero uncertainty in the momentumstarts with showing that the quantity

∫all spaceψ∗(x)(p^−p¯)2ψ(x)dx

Is (Δp)2. Then using the differential operator form ofp^and integration by parts, show that it is also,

∫all space{(p^−p¯)ψ(x)}∗{(p^−p¯)ψ(x)}dx

Together these show that if Δp is. 0. then the preceding quantity must be 0. However, the Integral of the complex square of a function(the quantity in the brackets) can only be 0 if the function is identically 0, so the assertion is proved.

Verify that solution (5-19) satisfies the Schrodinger equation in form (5.18).
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