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Modern Physics
Found in: Page 278
Modern Physics

Modern Physics

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

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Short Answer

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

The spherical harmonics Θl,mlθΦmlfare dimensionless.

All radial functions have dimension L-3/2.

The dimension of Pris 1L=L-1 .

See the step by step solution

Step by Step Solution

Step 1: Dimensional analysis

In engineering dimensional analysis is the analysis of relationship of physical quantities with each other by identifying their base quantities or the basic units.

Step 2:  Formula used

The normalization equation is given by,

0R2rr2dr=1

Where, R is the Radial wave function, r is the radius

Step 3:  The dimensions of Spherical Harmonics and radial functions

Consider table 7.3, the spherical harmonic functions represented as Θl,mlθΦmlfare the combinations of sine, cosine and complex exponential functions and thus they are dimension less.

As you can see in Table 7.4, all radial functions have dimension L-3/2, because their square gives probability per unit volume.

Step 4:  The dimensions of Pr

The dimension of Pr is calculated as,

0R2rr2dr=1

0P(r)dr=1

Where,Pr=r2R2r is the Radial Probability

Since, drhas the dimension of length so, the dimension of Pris1L=L-1 .

Most popular questions for Physics Textbooks

The Diatomic Molecule: Exercise 80 discusses the idea of reduced mass. Classically or quantum mechanically, we can digest the behavior of a two-particle system into motion of the center of mass and motion relative to the center of mass. Our interest here is the relative motion, which becomes a one-particle problem if we merely use μ for the mass for that particle. Given this simplification, the quantum-mechanical results we have learned go a long way toward describing the diatomic molecule. To a good approximation, the force between the bound atoms is like an ideal spring whose potential energy is 12kx2, where x is the deviation of the atomic separation from its equilibrium value, which we designate with an a. Thus,x=r-a . Because the force is always along the line connecting the two atoms, it is a central force, so the angular parts of the Schrödinger equation are exactly as for hydrogen, (a) In the remaining radial equation (7- 30), insert the potential energy 12kx2and replace the electron mass m with μ. Then, with the definition.f(r)=rR(r), show that it can be rewritten as

-ħ22μd2dr2f(r)+ħ2I(I+1)2μr2f(r)+12kx2f(r)=Ef(r)

With the further definition show that this becomes

-ħ22μd2dx2g(x)+ħ2I(I+1)2μ(x+a)g(x)+12kx2g(x)=Eg(x)

(b) Assume, as is quite often the case, that the deviation of the atoms from their equilibrium separation is very small compared to that separation—that is,x<<a. Show that your result from part (a) can be rearranged into a rather familiar- form, from which it follows that E=(n+12)ħkμ+ħ2I(I+1)2μa2 n=0,1,2,...I=0,1,2,...

(c)

Identify what each of the two terms represents physically.

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