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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 have three particles, and three distinct one-particle stateΨaX,ΨbX,andΨcxare available. How many different three-particle states can be constructed (a) if they are distinguishable particles, (b) if they are identical bosons, (c) if they are identical fermions? (The particles need not be in different states -ΨaX1,ΨaX2Ψax3would be one possibility, if the particles are distinguishable.)

Suppose you had three particles, one in stateψa(x), one in stateψb(x), and one in stateψc(x). Assuming ψa,ψb, andψc are orthonormal, construct the three-particle states (analogous to Equations 5.15,5.16, and 5.17) representing

(a) distinguishable particles,

(b) identical bosons, and

(c) identical fermions.

Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely antisymmetric, in the same sense. Comment: There's a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose first row isψa(x1),ψb(x1),ψc(x1) , etc., whese second row isψa(x2),ψb(x2),ψc(x2) , etc., and so on (this device works for any number of particles).

We can extend the theory of a free electron gas (Section 5.3.1) to the relativistic domain by replacing the classical kinetic energy, E=p2/2m,,with the relativistic formula, E=p2c2+m2c4-mc2. Momentum is related to the wave vector in the usual way: p=hk. In particular, in the extreme relativistic limit, E≈pc=hck.

(a) Replace h2k2n Equation 5.55 by the ultra-relativistic expression, hck, and calculateEtotin this regime.

dE=h2k22mVÏ€2k2dk (5.55).

(b) Repeat parts (a) and (b) of Problem 5.35 for the ultra-relativistic electron gas. Notice that in this case there is no stable minimum, regardless of R; if the total energy is positive, degeneracy forces exceed gravitational forces, and the star will expand, whereas if the total is negative, gravitational forces win out, and the star will collapse. Find the critical number of nucleons, Nc , such that gravitational collapse occurs for N>N_{C}is called the Chandrasekhar limit.

(c) At extremely high density, inverse beta decaye-+p+→n+v,converts virtually all of the protons and electrons into neutrons (liberating neutrinos, which carry off energy, in the process). Eventually neutron degeneracy pressure stabilizes the collapse, just as electron degeneracy does for the white dwarf (see Problem 5.35). Calculate the radius of a neutron star with the mass of the sun. Also calculate the (neutron) Fermi energy, and compare it to the rest energy of a neutron. Is it reasonable to treat a neutron star non relativistic ally?

(a) If ψaandψb are orthogonal, and both normalized, what is the constant A in Equation 5.10?

(b) Ifrole="math" localid="1658225858808" ψa=ψb (and it is normalized), what is A ? (This case, of course, occurs only for bosons.)

(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=8Ï€²¹3e-2r1+r2/a(5.30).

r1-r2=r12+r22-2r1r2³¦´Ç²õθ2.

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).

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