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If the constantCxinequation(7-5)were positive, the general mathematical solution would be

Ae+cxx+Be-cxx

Show that this function cannot be 0 at two points. This makes it an unacceptable solution for the infinite well, since it cannot be continuous with the wave functions outside the walls, which are 0.

Short Answer

Expert verified

This is possible only ifx1=x2 . (i.e., only one point) but this assumption does not happen in general cases. Thus, this is quite a contradiction to the assumption that the function is zero at two different pointsx1 and x2.

Step by step solution

01

Given data

The solution is:

Ae+cxx+Be-cxx

02

To determine the function Ae+cxx+Be-cxx cannot be zero at two points

The infinity well problem is one of the important problems in quantum mechanics that help us to understand other phenomena.

The wave function outside the well is zero.

The given solution is:

Ae+cxx+Be-cxx

It can also be written as:

Aexp+cxx+Bexp-cxx

These equations at two points x1,x2can be written as:

role="math" localid="1659763359765" Aexp+cxx1+Bexp-cxx1Aexp+cxx2+Bexp-cxx2

Solve each equation for B/A as:

role="math" localid="1659763407252" BA=exp2cxx1BA=exp2cxx2

From the above two expressions, we can conclude that,

exp2cxx2=exp2cxx1

03

Conclusion

This is possible only ifx1=x2 . (i.e., only one point) but this assumption does not happen in general cases. Thus, this is quite a contradiction to the assumption that the function is zero at two different pointsx1 and x2.

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Most popular questions from this chapter

An electron is in the 3d state of a hydrogen atom. The most probable distance of the electron from the proton is9ao. What is the probability that the electron would be found between8aoand10ao?

The space between two parallel metal plates is filled with an element in a gaseous stale. Electrons leave one plate at negligible speed and are accelerated toward the other by a potential differenceVapplied between the plates. As Vis increased from 0, the electron current increases more or less linearly, but when Vreaches 4.9 V , the current drops precipitously. From nearly 0 , it builds again roughly linearly as Vis increased beyond 4.9 V .

(a) How can the presence of the gas explain these observations?

(b) The Gas emits pure “light” when Vexceeds 4.9 V . What is its wavelength?

For the more circular orbits, =n-1and

P(r)r2ne-2r/na0

a) Show that the coefficient that normalizes this probability is

localid="1660047077408" (2na0)2n+11(2n)!

b) Show that the expectation value of the radius is given by

r¯=n(n+12)a0

and the uncertainty by

Δr=na0n2+14

c) What happens to the ratioΔr/r¯in the limit of large n? Is this large-n limit what would be expected classically?

For states where l = n - t the radial probability assumes the general form given in Exercise 54. The proportionality constant that normalizes this radial probability is given in Exercise 64.

(a) Show that the expectation value of the hydrogen atom potential energy is exactly twice the total energy. (It turns out that this holds no matter what l may be)

(b) Argue that the expectation value of the kinetic energy must be the negative of the total energy.

Some degeneracies are easy to understand on the basis of symmetry in the physical situation. Others are surprising, or “accidental”. In the states given in Table 7.1, which degeneracies, if any, would you call accidental and why?

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