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Find the lowest bound on the ground state of hydrogen you can get using a Gaussian trial wave function

(r)=Ae-br2,

where A is determined by normalization and b is an adjustable parameter. Answer-11.5eV

Short Answer

Expert verified

The lowest bound on the ground state of hydrogen using Gaussian trial wave function isHmin=-11.5eV.

Step by step solution

01

Step 1: Gaussian trail wave function

A Gaussian function is proposed as a trial wave function in a variational calculation on the hydrogen atom. Determine the optimum value of the parameter and the ground state energy of the hydrogen atom. Use atomic units

h=2蟿蟿,me=1,e=1

(r,):=(2)34exp(-r2)

Thegiventrial wave function is of the form:

(r)=Ae-br2

02

Finding the lowest bound on the ground state of hydrogen.

First, we find the normalization constant A:

2(r)r2sindrd=1A20e-2br2r2dr0xsind02xd=1A2182(b)3(2)(2)=1A=2b3/4

V=-e2400A240e-2br21rr2dr=-e24002b3/2414b=-e240022b.

03

Step 3: Finding the value of T

Now we find <T>

T=-h22mA2e-br22e-br2r2sindrddBut2e-br2=1r2ddrr2ddre-br2=1r2ddr-2br3e-br2=-2br2(3r2-2br4)e-br2=-h22m蟺产42b3/24(-2b)0(3r2-2br4)e-2br2dr=h2m蟺产42b3/2318b2b-2b332b22b.=h2m4蟺产2b38b-316b=3h2b2m.

04

Step 4: Finding the value of H

The last two calculations of the results to get <H>

H=3h2b2m-e24o022b;Hb=3h22m-e24o021b=0b=e24o022m3h2.Hmin=3h22me24o0224m29h2-e24o022e24o022m3h2=e24o02mh243-83.=-m2h2e24o0283=83E1=-11.5eV.

Thus the lowest bound on the ground state of hydrogen using Gaussian trial wave function isHmin=-11.5eV.

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Most popular questions from this chapter

Find the best bound on Egsfor the one-dimensional harmonic oscillator using a trial wave function of the form role="math" localid="1656044636654" (x)=Ax2+b2.,where A is determined by normalization and b is an adjustable parameter.

a) Use the variational principle to prove that first-order non-degenerate perturbation theory always overestimates (or at any rate never underestimates) the ground state energy.

(b) In view of (a), you would expect that the second-order correction to the ground state is always negative. Confirm that this is indeed the case, by examining Equation 6.15.

Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region.The 鈥渁rms鈥 of the cross continue out to infinity. The potential is zero within the cross, and infinite in the shaded areas outside. Surprisingly, this configuration admits a positive-energy bound state

(a) Show that the lowest energy that can propagate off to infinity is

Ethreshold=2h28ma2

any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say xa), and solve the Schr枚dinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the formexp(ikxx)withkx>0

(b) Now use the variation principle to show that the ground state has energy less than Ethreshold. Use the following trial wave function (suggested by Jim Mc Tavish):

(x,y)=A{cos(蟺虫/2a)+cos(蟺测/2a)e-xaandyacos(x/2a)e-y/axaandy>acos(y/2a)e-y/ax.aandya0elsewhere

Normalize it to determine A, and calculate the expectation value of H.
Answer:

<H>=h2ma2[28-1-(/4)1+(8/2)+(1/2)]

Now minimize with respect to 伪, and show that the result is less thanEthreshold. Hint: Take full advantage of the symmetry of the problem鈥 you only need to integrate over 1/8 of the open region, since the other seven integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not鈥攖here are 鈥渞oof-lines鈥 at the joins, and you will need to exploit the technique of Example 8.3.

Apply the techniques of this Section to the H-and Li+ions (each has two electrons, like helium, but nuclear charges Z=1and Z=3, respectively). Find the effective (partially shielded) nuclear charge, and determine the best upper bound on Egs, for each case. Comment: In the case of H- you should find that (H)>-13.6eV, which would appear to indicate that there is no bound state at all, since it would be energetically favourable for one electron to fly off, leaving behind a neutral hydrogen atom. This is not entirely surprising, since the electrons are less strongly attracted to the nucleus than they are in helium, and the electron repulsion tends to break the atom apart. However, it turns out to be incorrect. With a more sophisticated trial wave function (see Problem 7.18) it can be shown that Egs<-13.6eVand hence that a bound state does exist. It's only barely bound however, and there are no excited bound states, soH- has no discrete spectrum (all transitions are to and from the continuum). As a result, it is difficult to study in the laboratory, although it exists in great abundance on the surface of the sun.

Use a gaussian trial function (Equation 7.2) to obtain the lowest upper bound you can on the ground state energy of (a) the linear potential V(x)=|x| (b) the quartic potential:V(x)=x4

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