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(a) Find the percent error in Stirling’s approximation for z = 10 ?

(b)What is the smallest integer z such that the error is less than 1%?

Short Answer

Expert verified

(a) Error of Stirling's approximation for z =10 is 13.75%.

(b) Error of Stirling's approximation is less than 1% for z = 90.

Step by step solution

01

Define the Stirling’s approximation

  • Stirling's approximation (or Stirling's formula) is a factorial approximation in mathematics.
  • It is a good approximation, yielding accurate results even for low displaystyle nn values. It takes its name from James Stirling.
02

Calculating the percent error

(a)

For z = 10 we have

In(10!)=15.104410In(10)-10=13.0259%Error=15.1044-13.025915.104413.76%

Therefore the percent error in Stirling’s approximation is 13.76% .

03

Calculating the smallest integer

(b)

% Error is calculated by:

In(z!)zIn(z)+zIn(2!).100z%Error215.671000.89900.996851.06891.009

Therefore we see that for z = 90 and more error is less than 1%.

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Most popular questions from this chapter

Suppose you could find a solutionψ(r1,r2,...,rz)to the Schrödinger equation (Equation 5.25), for the Hamiltonian in Equation 5.24. Describe how you would construct from it a completely symmetric function, and a completely anti symmetric function, which also satisfy the Schrödinger equation, with the same energy.

role="math" localid="1658219144812" H^=j=1Z-ħ22mj2-14πo,0Ze2rj+1214πo,0j1Ze2rj-rk (5.24).

role="math" localid="1658219153183" H^ψ=E (5.25).

Find the energy at the bottom of the first allowed band, for the caseβ=10 , correct to three significant digits. For the sake of argument, assume αa=1eV.

Imagine two non interacting particles, each of mass , in the one dimensional harmonic oscillator potential (Equation 2.43). If one is in the ground state, and the other is in the first excited state, calculate (x1-x2)2assuming
(a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions. Ignore spin (if this bothers you, just assume they are both in the same spin state.)

(a) Write down the Hamiltonian for two noninteracting identical particles in the infinite square well. Verify that the fermion ground state given in Example 5.1 is an eigenfunction of H, with the appropriate eigenvalue.

(b) Find the next two excited states (beyond the ones in Example 5.1) - wave functions and energies - for each of the three cases (distinguishable, identical bosons, identical fermions).

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,and1=(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

-h22m112ψ-h22m222ψ+=-h22(m1+m2)R2ψ-h22μr2ψ+V(r)ψ=.(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.

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