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Chapter 5: Q26P (page 239)

Use the method of Lagrange multipliers to find the rectangle of largest area, with sides parallel to the axes that can be inscribed in the ellipse(xa)2+(yb)2=1. What is the maximum area?

Short Answer

Expert verified

The maximum area of an inscribed triangle can have is=2ab

Step by step solution

01

Define area of triangle

The area of a triangle is defined as the total space occupied by a triangle's three sides in a two-dimensional plane. The basic formula for calculating the area of a triangle is half the product of its base and height.

02

Calculating the area of inscribed rectangle

We want area of the inscribed rectangle to be maximum. If its sides are equal to 2x and 2y, its area is: a(x,y)=4xy.

G(x,y,λ)=4xy+λ(xa)2+(yb)2-1∂G∂x=0=4y+2λxa2y=-λx2a2∂G∂y=0=4x+2λyb2⇒4x=-λy2b2=-λ2xa2b2

λ1=0orλ2=±2ab

03

Calculating the maximum area

Because side of a rectangular must be positive

y=-λx2a2=±bxa=bxa∂G∂λ=0=(xa)2+(yb)2-1x2a2+b2x2a2b2=1x2=a22x=a2,y=a2

So maximum area of an inscribed triangle can have is A=4xy=2ab

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Most popular questions from this chapter

In view ofProblem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass.

(a) Find (to two significant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of μ.

E1=-m2h2e24π∞'2=-13.6eV(4.77).

(b) Find the separation in wavelength between the red Balmer lines n=3→n=2for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton).

(c) Find the binding energy of positronium (in which the proton is replaced by a positron—positrons have the same mass as electrons, but opposite charge).

(d) Suppose you wanted to confirm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the “Lyman-α” line n=2→n=1?.

(a) Calculate<1/r1-r2>for the stateψ0(Equation 5.30). Hint: Dod3r2integral

first, using spherical coordinates, and setting the polar axis alongr1, so

that

ψ0r1,r2=ψ100r1ψ100r2=8πa3e-2r1+r2/a(5.30).

r1-r2=r12+r22-2r1r2cosθ2.

Theθ2integral is easy, but be careful to take the positive root. You’ll have to

break ther2integral into two pieces, one ranging from 0 tor1,the other fromr1to∞

Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it toE0(Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we’re still working with an approximate wave function, so don’t expect perfect agreement.)

E0=8-13.6eV=-109eV(5.31).

a) Hund’s first rule says that, consistent with the Pauli principle, the state with the highest total spin (S) will have the lowest energy. What would this predict in the case of the excited states of helium?

(b) Hund’s second rule says that, for a given spin, the state with the highest total orbital angular momentum (L) , consistent with overall antisymmetrization, will have the lowest energy. Why doesn’t carbon haveL=2? Note that the “top of the ladder”(ML=L)is symmetric.

(c) Hund’s third rule says that if a subshell(n,l)is no more than half filled,
then the lowest energy level hasJ=lL-SI; if it is more than half filled, thenJ=L+Shas the lowest energy. Use this to resolve the boron ambiguity inProblem 5.12(b).

(d) Use Hund’s rules, together with the fact that a symmetric spin state must go with an antisymmetric position state (and vice versa) to resolve the carbon and nitrogen ambiguities in Problem 5.12(b). Hint: Always go to the “top of the ladder” to figure out the symmetry of a state.

Find the average energy per free electron (Etot/Nd), as a fraction of the

Fermi energy. Answer:(3/5)EF

Typically, the interaction potential depends only on the vectorr=r1-r2between

the two particles. In that case the Schrodinger equation seperates, if we change variables from r1,r2andR=(m1r1+m2r2)I(m1+m2)(the center of mass).

(a)Show that

localid="1655976113066" r1=R+(μ/m1)r,r2=R-(μ/m)r,and∇1=(μ/m)∇R+∇r,∇2=(μ/m1)∇R-∇r,where

localid="1655976264171" μ=m1m2m1+m2(5.15).

is the reduced mass of the system

(b) Show that the (time-independent) Schrödinger equation (5.7) becomes

-h22m1∇12ψ-h22m2∇22ψ+Vψ=Eψ-h22(m1+m2)∇R2ψ-h22μ∇r2ψ+V(r)ψ=Eψ.(5.7).

(c) Separate the variables, letting ψ(R,r)=ψR(R)ψr(r)Note that ψRsatisfies the

one-particle Schrödinger equation, with the total mass (m1+m2)in place of m, potential zero, and energy ERwhile ψrsatisfies the one-particle Schrödinger equation with the reduced mass in place of m, potential V(r) , and energy localid="1655977092786" Er. The total energy is the sum: E=ER+Er. What this tells us is that the center of mass moves like a free particle, and the relative motion (that is, the motion of particle 2 with respect to particle 1) is the same as if we had a single particle with the reduced mass, subject to the potential V. Exactly the same decomposition occurs in classical mechanics; it reduces the two-body problem to an equivalent one-body problem.
See all solutions

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