Q#39.95
(a) Show that in the Bohr model, the frequency of revolution of an electron in its circular orbit around a stationary hydrogen nucleus is f = $\frac{me^4}{4\epsilon_0^2n^3h^3}$.
(b) In classical physics, the frequency of revolution of the electron is equal to the frequency of the radiation that it emits. Show that when n is very large, the frequency of revolution does indeed equal the radiated frequency calculated from Eq. (39.5) for a transition from $n_1$ = n + 1 to $n_2$ = n. (This illustrates Bohr’s correspondence principle, which is often used as a check on quantum calculations. When n is small, quantum physics gives results that are very different from those of classical physics. When n is large, the differences are not significant, and the two methods then “correspond.” In fact, when Bohr first tackled the hydrogen atom problem, he sought to determine ƒ as a function of n such that it would correspond to classical results for large n.)
Answer:
The period was found in Exercise 39.29b:
T = $\frac{4\epsilon_0^2n^3h^3}{me^2}$
Eq. (39.14) gives the energy of state n of a hydrogen atom.
(a) The frequency is f = $\frac{1}{T}=\frac{me^4}{4\epsilon_0^2n^3h^3}$
(b) From Eq. f = $\frac{1}{h}(E_2-E_1)$ and
the total energy is the sum of the kinetic and potential energies:
$E_n=-\frac{1}{\epsilon_0^2}\frac{me^4}{8n^2h^2}$
$E_2-E_1=\frac{1}{\epsilon_0^2}\frac{me^4}{8h^2}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$
If $n_1$ = n + 1 and $n_2$ = n,
$E_2-E_1=\frac{1}{\epsilon_0^2}\frac{me^4}{8h^2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)$
$E_2-E_1=\frac{1}{\epsilon_0^2}\frac{me^4}{8h^2}\frac{1}{n^2}\left(1-\frac{1}{(1+1/n)^2}\right)$
$E_2-E_1 \approx \frac{1}{\epsilon_0^2}\frac{me^4}{8h^2}\frac{1}{n^2}\left(1-\left(1-\frac{2}{n}+...\right)\right)$
$E_2-E_1 \approx \frac{1}{\epsilon_0^2}\frac{me^4}{8h^2}\frac{1}{n^3}$
because $1-\left(1-\frac{2}{n}+...\right)\approx \frac{2}{n}$
f = $\frac{1}{h}(E_2-E_1)$
$f \approx \frac{1}{\epsilon_0^2}\frac{me^4}{8h^3}\frac{1}{n^3}$
$f \approx \frac{me^4}{4\epsilon_0^2h^3n^3}$
We have shown that for large n we obtain the classical result that the frequency of revolution of the electron is equal to the frequency of the radiation it emits.
Q#39.96
You have entered a contest in which the contestants drop a marble with mass 20.0 g from the roof of a building onto a small target 25.0 m below. From uncertainty considerations, what is the typical distance by which you will miss the target, given that you aim with the highest possible precision? (Hint: The uncertainty $\Delta x_f$ in the x-coordinate of the marble when it reaches the ground comes in part from the uncertainty $\Delta x_i$ in the x-coordinate initially and in part from the initial uncertainty in $v_x$. The latter gives rise to an uncertainty $\Delta v_x$ in the horizontal motion of the marble as it falls. The values of $\Delta x_i$ and $\Delta v_x$ are related by the uncertainty principle. A small $\Delta x_i$ gives rise to a large $\Delta v_x$ and vice versa. Find the value $\Delta x_i$ of that gives the smallest total uncertainty in x at the ground. Ignore any effects of air resistance.)
Answer:
The value of $\Delta x_i$ that minimizes $\Delta x_f$ satisfies
$\frac{d\Delta x_f}{d\Delta x_i}=0$
Time of flight of the marble, from a free-fall kinematic equation is just
$t =\sqrt{\frac{2y}{g}}$
$t =\sqrt{\frac{2(25.0 \ m)}{9.8 \ m/s^2}}=2.26 \ s$
$\Delta x_f=\Delta x_i+(\Delta v_x)t$
because $\Delta p_x=m\Delta v_x$ and $\Delta p_x \Delta x \geq \frac{\hbar}{2}$
$\Delta x_f=\Delta x_i+\left(\frac{\Delta p_x}{m}\right)t$ or
$\Delta x_f=\Delta x_i+\left(\frac{\hbar t}{2 \Delta x_i m}\right) + \Delta x_i$
To minimize $\Delta x_f$ to $\Delta x_i$,
$\frac{d\Delta x_f}{d\Delta x_i}=0$
$\left(\frac{-\hbar t}{2 (\Delta x_i)^2 m}\right) + 1=0$
$\frac{-\hbar t}{2 (\Delta x_i)^2 m} = -1$
$\Delta x_i^2=\frac{\hbar t}{2 m} $
$\Delta x_i(min)=\sqrt{\frac{\hbar t}{2 m}} $ and
$\Delta x_f(min)=\sqrt{\frac{\hbar t}{2 m}} +\sqrt{\frac{\hbar t}{2 m}} = \sqrt{\frac{2\hbar t}{ m}}$
$\Delta x_f(min)=\sqrt{\frac{2(1.055 \times 10^{-34} \ J.s)(2.26 \ s)}{0.0200 \ kg}} $
$\Delta x_f(min)=1.54 \times 10^{-16} \ m = 1.54 \times 10^{-7} \ nm$
The uncertainty introduced by the uncertainty principle is completely negligible in this situation.
Post a Comment for "Particles Behaving as Waves Problems and Solutions - CHALLENGE PROBLEMS"