91影视

An electron in a hydrogen atom is in the (n,l,m_l) = (2,1,0) state.

(a) Calculate the probability that it would be found within 60 degrees of z-axis, irrespective of radius.

(b) Calculate the probability that it would be found between r = 2a₀ and r = 6a₀, irrespective of angle.

(c) What is the probability that it would be found within 60 degrees of the z-axis and between r = 2a₀ and r = 6a₀?

Show that a transition where$\mathbf{鈭�}{\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{卤}\mathbf{1}$corresponds to a dipole moment in the xy-plane, while $\mathbf{鈭�}{\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$ corresponds to a moment along the z-axis. (You need to consider only the $\mathit{蠒}$ -parts of therole="math" localid="1659783155213" ${\mathit{蠄}}_i$ and${\mathit{蠄}}_{\mathbf{f}}$ , which are of the form ${\mathit{e}}^{\mathbf{i}{\mathbf{m}}_{\mathbf{l}}\mathbf{蠒}}$):

Verify for the angular solutions $鈯�\left(胃\right)蠁\left(蠒\right)$of Table 7.3 that replacing $蠒$ with $蠒+\mathrm{蟺}$ and replacing $胃$ with $蟺-胃$gives the same function whenis even and the negative of the function when lis odd.

A particular vibrating diatomic molecule may be treated as a simple harmonic oscillator. Show that a transition from that n=2state directly to n=0ground state cannot occur by electric dipole radiation.

Consider a vibrating molecule that behaves as a simple harmonic oscillator of mass ${\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$, spring constant 10³N/m and charge is +e , (a) Estimate the transition time from the first excited state to the ground state, assuming that it decays by electric dipole radiation. (b) What is the wavelength of the photon emitted?

Calculate the electric dipole moment p and estimate the transition time for a hydrogen atom electron making an electric dipole transition from the to the $\mathbf{(}\mathbf{n}\mathbf{,}\mathbf{l}\mathbf{,}\mathbf{m}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{+}\mathbf{1}\mathbf{)}$ ground state. Comment on the relationship of the result to that in Example 7.11.

Calculate the electric dipole moment p and estimate the transition time for a hydrogen atom electron making an electric dipole transition from the

$\left(n,l,m\right)\mathbf{=}\left(3,2,0\right)$to the $\left(2,1,0\right)$ state.

When applying quantum mechanics, we often concentrate on states that qualify as 鈥渙rthonormal鈥�, The main point is this. If we evaluate a probability integral over all space of ${{\mathbf{蠒}}_{\mathbf{1}}}^{\mathbf{*}}{\mathit{蠒}}_{\mathbf{1}}$or of ${{\mathbf{蠒}}_{\mathbf{2}}}^{\mathbf{*}}{\mathit{蠒}}_{\mathbf{2}}$, we get 1 (unsurprisingly), but if we evaluate such an integral for${{\mathbf{蠒}}_{\mathbf{1}}}^{\mathbf{*}}{\mathit{蠒}}_{\mathbf{2}}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{}{{\mathbf{蠒}}_{\mathbf{2}}}^{\mathbf{*}}{\mathit{蠒}}_{\mathbf{1}}$ we get 0. This happens to be true for all systems where we have tabulated or actually derived sets of wave functions (e.g., the particle in a box, the harmonic oscillator, and the hydrogen atom). By integrating overall space, show that expression (7-44) is not normalized unless a factor of $\mathbf{1}\mathbf{/}\mathbf{2}$is included with the probability.

Mathematically equation (7-22) is the same differential equation as we had for a particle in a box-the function and its second derivative are proportional. But$\mathbf{\left(}\mathit{蠒}\mathbf{\right)}$for m₁= 0is a constant and is allowed, whereas such a constant wave function is not allowed for a particle in a box. What physics accounts for this difference?

Consider two particles that experience a mutual force but no external forces. The classical equation of motion for particle 1 is ${\stackrel{\mathbf{藱}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{=}{\mathbf{F}}_{\mathbf{2}\mathbf{on}\mathbf{1}}\mathbf{/}{\mathbf{m}}_{\mathbf{1}}$, and for particle 2 is ${\stackrel{\mathbf{藱}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}{\mathbf{F}}_{\mathbf{1}\mathbf{on}\mathbf{2}}\mathbf{/}{\mathbf{m}}_{\mathbf{2}}$, where the dot means a time derivative. Show that these are equivalent to ${\stackrel{\mathbf{藱}}{\mathbf{v}}}_{\mathbf{cm}}\mathbf{=}\mathbf{constant}$, and ${\stackrel{\mathbf{藱}}{\mathbf{v}}}_{\mathbf{rel}}\mathbf{=}{\mathbf{F}}_{\mathbf{Mutual}}\mathbf{/}\mathbf{渭}$ .Where, ${\stackrel{\mathbf{藱}}{\mathbf{v}}}_{\mathbf{cm}}\mathbf{=}\left(m_1{\stackrel{藱}{v}}_1+m_2{\stackrel{藱}{v}}_2\right)\mathbf{/}\left(m_1{m}_2\right)\mathbf{,}{\mathbf{F}}_{\mathbf{Mutual}}\mathbf{=}\mathbf{-}{\mathbf{F}}_{\mathbf{ion}\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{渭}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{\left(m_1+m_2\right)}$.

In other words, the motion can be analyzed into two pieces the center of mass motion, at constant velocity and the relative motion, but in terms of a one-particle equation where that particle experiences the mutual force and has the 鈥渞educed mass鈥� $\mathbf{渭}$.