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Chapter 5: 6CQ (page 186)

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

Short Answer

Expert verified

Answer:

The likelihood of discovering the probability would be 0 if the temporal part of a wave function was zero, implying that the particle would have vanished.

Step by step solution

01

Theory of wave function

The wave function $ψ$ itself has no physical significance. However, the amplitude of wave function corresponds $ψ$ to and the square of the wave function $ψ$ relates to the photon density, the number of photons present in a region, it relates to electron density in a certain region.

02

Conclusion

Therefore, using the square of wave function, we can measure of the probability that the electron can be found within a particular tiny volume of the atom.

Hence, The likelihood of discovering the probability would be 0 if the temporal part of a wave function was zero, implying that the particle would have vanished.

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Most popular questions from this chapter

A bound particle of massdescribed by the wave function

${ψ(x)=Axe}^{{-x}^{2} {/2b}^{2}}$

What is the most probable location at which to find the particle?

The deeper the finite well, the more state it holds. In fact, a new state, the, is added when the well’s depth $U_{0}$ reaches $h^{2} {(n - 1)}^{2} / 8 m L^{2}$ . (a) Argue that this should be the case based only on $k = \sqrt{2 m E / h^{2}}$ , the shape of the wave inside, and the degree of penetration of the classically forbidden region expected for a state whose energy E is only negligibly below $U_{0}$ . (b) How many states would be found up to this same “height” in an infinite well.

Whereas an infinite well has an infinite number of bound states, a finite well does not. By relating the well height $U_{0}$ to the kinetic energy and the kinetic energy (through $λ$ ) to n and L. Show that the number of bound states is given roughly by $\sqrt{8 m l^{2} U_{0} / h^{2}}$ (Assume that the number is large.)

A comet in an extremely elliptical orbit about a star has, of course, a maximum orbit radius. By comparison, its minimum orbit radius may be nearly 0. Make plots of the potential energy and a plausible total energyversus radius on the same set of axes. Identify the classical turning points on your plot.

Figure 5.15 shows that the allowed wave functions for a finite well whose depth $U_{0}$ was chosen to be $6 π ℏ^{2} / m L^{2}$ .

(a) Insert this value in equation (5-23), then using a calculator or computer, solve for the allowed value of $k L$ , of which there are four.

(b) Using $k = \frac{\sqrt{2 m E}}{ℏ}$ find corresponding values of E. Do they appear to agree with figure 5.15?

(c) Show that the chosen $U_{0}$ implies that $α ≃ \sqrt{\frac{12 π^{2}}{L^{2}} - k^{2}}$ .

(d) Defining $L$ and $C$ to be 1 for convenience, plug your $K L$ and $α$ values into the wave function given in exercise 46, then plot the results. ( Note: Your first and third $K L$ values should correspond to even function of z, thus using the form with $C O S K Z$ , while the second and forth correspond to odd functions. Do the plots also agree with Figure 5.15?

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