Using the functions given in Table 7.4, verify that for the more circular electron orbit in hydrogen (i.e.,l=n-1), the radial probability is of the form

P(r)r2ne-2r/nao

Show that the most probable radius is given by

rmostprobable=n2ao

Short Answer

Expert verified

The most probable radius isr=n2a0.

Step by step solution

01

 Given data

The radial function for l=n-1is,

role="math" localid="1659182273339" Rn,n-1rn-1e-r/(nao)

02

 Concept

The most probable radius is the radius of the orbit in which the probability of finding the electron is maximum.

The probability density is

P(r)=r2R2n,n-1

03

 Solution

P(r)=r2R2n,n-1=r2rn-1e-r/(nao)=r2r2n-2e-r/(nao)=r2ne-r/(nao)

To find the most probable location, take the derivative of the probability density,

dPrdr=ddrr2ne-2rlna°=2nr2n-1e-2rlna°+r2n-2na°e-2rlna°=e-2rlna°r2n-12n+-2rna°

For the most probable location, the derivative should be equal to zero.

role="math" localid="1659182748704" =e-2rlna°r2n-12n+-2rna°

Hence,

2n+-2rna°=0r=n2a°

The most probable radius is r=n2a°.

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Most popular questions from this chapter

What are the dimensions of the spherical harmonics Θl,ml(θ)Φml(ϕ)given in Table 7.3? What are the dimensions of theRn,l(r)given in Table 7.4, and why? What are the dimensions ofP(r), and why?

Verify that the solution given in equation (7.6) satisfy differential equations (7.5) as well as the required boundary conditions.

In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,λ=2L/n , where is the length of its home. Given that λ=h/p, we would have p=nh/2L, and the kinetic energy, p2/2m, would thus be n2h2/8mL2. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: E=U. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?

Consider a 2D infinite well whose sides are of unequal length.

(a) Sketch the probability density as density of shading for the ground state.

(b) There are two likely choices for the next lowest energy. Sketch the probability density and explain how you know that this must be the next lowest energy. (Focus on the qualitative idea, avoiding unnecessary reference to calculations.)

A mathematical solution of the azimuthal equation (7-22) is Φ(φ)=Ae-+Be- , which applies when D is negative, (a) Show that this simply cannot meet itself smoothly when it finishes a round trip about the z-axis. The simplest approach is to consider φ=0 and φ=2π. (b) If D were 0, equation (7-22) would say simply that the second derivative Φ(φ)of is 0. Argue than this too leads to physically unacceptable solution, except in the special case of Φ(φ) being constant, which is covered by the ml=0 , case of solutions (7-24).

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