91影视

Exercise 80 discusses the idea of reduced mass. When two objects move under the influence of their mutual force alone, we can treat the relative motion as a one particle system of mass $\mathbf{渭}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{2}}\mathbf{/}\left(m_1+m_2\right)$. Among other things, this allows us to account for the fact that the nucleus in a hydrogen like atom isn鈥檛 perfectly stationary, but in fact also orbits the centre of mass. Suppose that due to Coulomb attraction, an object of mass ${\mathbf{m}}_{\mathbf{2}}$and charge $\mathbf{-}\mathbf{e}$orbits an object of mass ${\mathbf{m}}_{\mathbf{1}}$ and charge +Ze . By appropriate substitution into formulas given in the chapter, show that (a) the allowed energies are $\frac{{\mathbf{Z}}^{\mathbf{2}}\mathbf{渭}}{\mathbf{m}\mathbf{}}\frac{{\mathbf{E}}_{\mathbf{1}}}{{\mathbf{n}}^{\mathbf{2}}}$, where is the hydrogen ground state, and (b) the 鈥淏ohr Radius鈥� for this system is $\frac{\mathbf{m}}{\mathbf{锄渭}}{\mathbf{a}}_{\mathbf{0}}$ ,where ${\mathbf{a}}_{\mathbf{0}}$is the hydrogen Bohr radius.

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass, ${\mathit{m}}_{\mathbf{1}}$while ${\mathit{m}}_2$is the mass of the orbiting negative charge. (a) What percentage error is introduced in the hydrogen ground-state energy by assuming that the proton is of infinite mass? (b) Deuterium is a form of hydrogen in which a neutron joins the proton in the nucleus, making the nucleus twice as massive. Taking nuclear mass into account, by what percent do the ground-state energies of hydrogen and deuterium differ?

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass ${\mathit{m}}_{\mathbf{1}}$, while${\mathit{m}}_{\mathbf{2}}$is the mass of the orbiting negative charge. In positronium, an electron orbits a single positive charge, as in hydrogen, but one whose mass is the same as that of the electron -- a positron. Obtain numerical values of the ground state energy and 鈥淏ohr radius鈥� of positronium.

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass$m_1$, while$m_2$is the mass of the orbiting negative charge. In positronium, an electron orbits a single positive charge, as in hydrogen, but one whose mass is the same as that of the electron -- a positron. Obtain numerical values of the ground state energy and 鈥淏ohr radius鈥� of positronium.

Classically, it was expected that an orbiting electron would emit radiation of the same frequency as its orbit frequency. We have often noted that classical behaviour is observed in the limit of large quantum numbers. Does it work in this case? (a) Show that the photon energy for the smallest possible energy jump at the 鈥渓ow-n-end鈥� of the hydrogen energies is $\mathbf{3}\mathbf{\left|}{\mathbf{E}}_{\mathbf{0}}\mathbf{\right|}\mathbf{/}{\mathbf{n}}^{\mathbf{3}}$, while that for the smallest jump at the 鈥渉igh-n-end鈥� is $\mathbf{2}\mathbf{\left|}{\mathbf{E}}_{\mathbf{0}}\mathbf{\right|}\mathbf{/}{\mathbf{n}}^{\mathbf{3}}$, where ${\mathbf{E}}_{\mathbf{0}}$is hydrogen鈥檚 ground-state energy. (b) Use F=ma to show that the angular velocity of a classical point charge held in orbit about a fixed-point charge by the coulomb force is given by $\mathbf{蝇}\mathbf{=}\sqrt{{\mathbf{e}}^{\mathbf{2}}\mathbf{/}\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}{\mathbf{mr}}^{\mathbf{3}}}$. (c) Given that $\mathbf{r}\mathbf{=}{\mathbf{n}}^{\mathbf{2}}{\mathbf{a}}_{\mathbf{0}}$, is this angular frequency equal to the minimum jump photon frequency at either end of hydrogen鈥檚 allowed energies?

The expectation value of the electron鈥檚 kinetic energy in the hydrogen ground state equals the magnitude of the total energy (see Exercise 60). What must be the width of a cubic finite wall, in terms of $a_0$, for this ground state to have this same energy?

Spectral lines are fuzzy due to two effects: Doppler broadening and the uncertainty principle. The relative variation in wavelength due to the first effect (see Exercise 2.57) is given by

$\frac{\mathbf{鈭�}\mathbf{位}}{\mathbf{位}}\mathbf{=}\frac{\sqrt{\mathbf{3}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}\mathbf{/}\mathbf{m}}}{\mathbf{c}}$

Where T is the temperature of the sample and m is the mass of the particles emitting the light. The variation due to the second effect (see Exercise 4.72) is given by

$\frac{\mathbf{鈭�}\mathbf{位}}{\mathbf{位}}\mathbf{=}\frac{\mathbf{位}}{\mathbf{4}\mathbf{蟺肠}}$

Where, $\mathbf{鈭�}\mathbf{t}$ is the typical transition time

(a) Suppose the hydrogen in a star has a temperature of $\mathbf{5}\mathbf{脳}{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{K}$. Compare the broadening of these two effects for the first line in the Balmer series (i.e.,${\mathbf{n}}_{\mathbf{i}}\mathbf{=}\mathbf{3}\mathbf{鈫�}{\mathbf{n}}_{\mathbf{f}}\mathbf{=}\mathbf{2}$ ). Assume a transition time of 10^-8s. Which effect is more important?

(b) Under what condition(s) might the other effect predominate?

A comet of ${\mathbf{10}}^{\mathbf{14}}\mathbf{\text{鈥�}}\mathbf{kg}$ mass describes a very elliptical orbit about a star of mass$\mathbf{3}\mathbf{脳}{\mathbf{10}}^{\mathbf{30}}\mathbf{\text{鈥�}}\mathbf{kg}$ , with its minimum orbit radius, known as perihelion, being role="math" localid="1660116418480" ${\mathbf{10}}^{\mathbf{11}}\mathit{m}$ and its maximum, or aphelion, $\mathbf{100}$ times as far. When at these minimum and maximum

radii, its radius is, of course, not changing, so its radial kinetic energy is $\mathbf{0}$, and its kinetic energy is entirely rotational. From classical mechanics, rotational energy is given by $\frac{{\mathbf{L}}^{\mathbf{2}}}{\mathbf{2}\mathbf{I}}$, where $\mathit{I}$is the moment of inertia, which for a 鈥減oint comet鈥� is simply ${\mathbf{mr}}^{\mathbf{2}}$.

(a) The comet鈥檚 speed at perihelion is$\mathbf{6}\mathbf{.}\mathbf{2945}\mathbf{脳}{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{鈥�}}\mathbf{m}\mathbf{/}\mathbf{s}$ . Calculate its angular momentum.

(b) Verify that the sum of the gravitational potential energy and rotational energy are equal at perihelion and aphelion. (Remember: Angular momentum is conserved.)

(c) Calculate the sum of the gravitational potential energy and rotational energy when the orbit radius is $\mathbf{50}$ times perihelion. How do you reconcile your answer with energy conservation?

(d) If the comet had the same total energy but described a circular orbit, at what radius would it orbit, and how would its angular momentum compare with the value of part (a)?

(e) Relate your observations to the division of kinetic energy in hydrogen electron orbits of the same $\mathbf{n}$but different $\mathbf{I}$.

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 $\mathbf{渭}$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 $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$, 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 Schr枚dinger equation are exactly as for hydrogen, (a) In the remaining radial equation (7- 30), insert the potential energy $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$and replace the electron massm with $\mathbf{渭}$. Then, with the definition.$\mathbf{f}\left(r\right)\mathbf{=}\mathbf{rR}\left(r\right)$, show that it can be rewritten as

$\frac{\mathbf{-}{\mathbf{魔}}^{\mathbf{2}}}{\mathbf{2}\mathbf{渭}}\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathit{f}\left(r\right)\mathbf{+}\frac{{\mathbf{魔}}^{\mathbf{2}}\mathbf{I}\left(I+1\right)}{\mathbf{2}\mathbf{渭}{\mathbf{r}}^{\mathbf{2}}}\mathit{f}\left(r\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}\mathit{f}\left(r\right)\mathbf{=}\mathit{E}\mathit{f}\left(r\right)$

With the further definition show that this becomes

$\frac{\mathbf{-}{\mathbf{魔}}^{\mathbf{2}}}{\mathbf{2}\mathbf{渭}}\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{魔}}^{\mathbf{2}}\mathbf{I}\mathbf{(}\mathbf{I}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}\mathbf{渭}\left(x+a\right)}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{E}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

(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鈥攖hat is,x<<a. Show that your result from part (a) can be rearranged into a rather familiar- form, from which it follows that
$\mathit{E}\mathbf{=}\left(n+\frac{1}{2}\right)\mathit{魔}\sqrt{\frac{\mathbf{k}}{\mathbf{渭}}}\mathbf{+}\frac{{\mathbf{魔}}^{\mathbf{2}}\mathbf{I}\left(I+1\right)}{\mathbf{2}\mathbf{渭}{\mathbf{a}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\begin{array}{c}\mathbf{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\\ \mathbf{I}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\end{array}$

(c)

Identify what each of the two terms represents physically.

A spherical infinite well has potential energy

$\mathit{U}\left(r\right)\mathbf{=}\{\begin{array}{l}0r<a\\ +鈭�r>a\end{array}$

Since this is a central force, we may use the Schrodinger equation in the form (7-30)-that is, just before the specific hydrogen atom potential energy is inserted. Show that the following is a solution

$\mathit{R}\left(r\right)\mathbf{=}\frac{\mathbf{A}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{b}\mathbf{r}}{\mathbf{r}}$

Now apply the appropriate boundary conditions. and in so doing, find the allowed angular momenta and energies for solutions of this form.