Съдържание

Условие Отговори Решение

USAPhO 1997, Полуфинал — Задача 4. A4

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Всички материали от USAPhO 1997

United States Physics Team — 1997 Exam 2

Указания

Entia non multiplicanda sunt praeter necessitatem

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Please show all work, and do not hesitate to use explanatory words in addition to equations. Partial credit will be awarded.
Begin each problem on a fresh sheet of paper. In the upper right-hand corner of each page, write your name, problem number, and the page number/total number of pages for this problem (e.g., "A2, 3/4" for "Problem A2, page 3 of four").
You may use a non-programmable calculator, but may not use any tables, books, notes, or collections of formulas.
Work Part A first. You have ninety minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.

Possibly useful information:

Gravitational field on Earth’s surface: $g = 9.8\ \mathrm{N/kg}$

Gravitational constant: $G = 6.67\times10^{-11}\ \mathrm{N\text{-}m^2 / kg^2}$

Coulomb constant: $k = 1 / 4\pi\epsilon_o = 8.99\times10^{9}\ \mathrm{N\text{-}m^2 / C^2}$

Speed of light in vacuum: $c = 3.00\times10^{8}\ \mathrm{m/s}$

Permeability of empty space: $\mu_o$ , where $\mu_o\epsilon_o = 1/c^2$

Ideal gas constant $R = 8.31\ \mathrm{J / mol\text{-}K}$

Boltzmann’s constant: $k_B = 1.38\times10^{-23}\ \mathrm{J/K}$

Avagadro’s number: $N_A = 6.02\times10^{23}$

1 electron volt = $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$

Planck’s constant: $h = 6.63\times10^{-34}\ \mathrm{J\text{-}s} = 4.14\times10^{-15}\ \mathrm{eV\text{-}s}$

$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \ldots$ for $|x| < 1$ $\qquad$ $\cos\theta \approx 1 - \theta^2/2$ for $\theta \ll 1$

$(1 + x)^n = 1 + nx + (1/2!)n(n-1)x^2 + \ldots$ for $|x| < 1$ $\qquad$ $\sin\theta \approx \theta$ for $\theta \ll 1$

Условие

An atom in an excited state typically decays by emitting a photon. The atom is originally in some excited state $S'$ , and after the decay it is in the ground state $S$ . We usually think of the photon as carrying away the energy difference between states $S'$ and $S$ . In this problem we examine this process closely.

(a) Consider a stationary, isolated atom in some excited state $S'$ . Before decay the mass of the atom is $M'$ . The atom decays into the ground state $S$ . In the ground state the mass of the atom is $M$ . Thus the energy $(M'-M)c^2$ , which we denote here as $E_o$ , is the energy that is released by the atom, and is therefore an upper limit on the energy that the photon could possibly carry away ( $c$ is the speed of light). Let $E$ denote the energy of the outgoing photon. Without solving any equation, by merely giving an qualitative argument, explain why $E$ must be strictly less than $E_o$ in the case of an isolated excited atom undergoing radiative decay. [4 т.]

(b) Quantify your answer to part (a) by showing that, approximately,

$E \approx E_o\,[\,1 - (\,E_o / 2Mc^2\,)\,]$ , where $E_o / 2Mc^2 \ll 1$ . [6 т.]

(c) Consider an atom decaying from the excited state $S'$ to the ground state $S$ , where $Mc^2 = 2\times10^{11}$ eV and $E_o = 4\times10^{5}$ eV. A photon energy corresponds to a radiation frequency. Thus the energy difference $E_o - E$ corresponds to a frequency shift $\Delta f = f_o - f$ . Calculate the fractional frequency shift $\Delta f / f_o$ . [5 т.]

(d) Another way to produce a frequency shift is the Doppler effect, produced by relative motion between source and observer. The atom of part (c) emits the photon energy $E$ , which corresponds to light frequency $f$ . With what speed and in what direction would an observer have to move relative to this photon, to Doppler-shift its frequency back to the value $f_o$ corresponding to the energy $E_o$ ? The observer’s speed $v$ is $\ll c$ . [5 т.]

Отговори

Решение

(a) To conserve momentum the atom must recoil. Thus after emitting the photon, the atom carries some kinetic energy $K$ , leaving less than $E_o$ for photon.

(b) Before decay: $(\text{Energy, momentum}) = (M'c^2, 0)$ . After decay: atom $(Mc^2 + K, -p)$ , photon $(E, E/c)$ .

Conservation of Energy

M'c^2 = Mc^2 + K + E

(M' - M)c^2 = K + E

E_o = K + E

Cons. of Momentum

0 = \frac{E}{c} - p

p = \frac{E}{c}

[since the atom is heavy, $v \ll c$ and $K = p^2/2m$ , or $p \approx \sqrt{2mK}$ for atom]

\therefore\ \sqrt{2mK} = \frac{E}{c} \ \Rightarrow\ K = \frac{E^2}{2mc^2}

$\{$ or, relativistically,

(Mc^2)^2 = (K + Mc^2)^2 - (pc)^2

\Rightarrow\ K = Mc^2\left[\sqrt{1 + \left(\frac{E}{Mc^2}\right)^2} - 1\right] \approx E^2/2mc^2 \ \}

[given $\dfrac{E_o}{2mc^2} \ll 1$ since $K$ and $E$ are $< E_o$ , $\dfrac{K}{2mc^2}$ and $\dfrac{E}{2mc^2}$ are $\ll 1$ too]

Conservation of Energy Equation becomes

E_o = \frac{E^2}{2mc^2} + E

\begin{aligned} \Rightarrow\ E &= E_o\left(1 + \frac{E}{2mc^2}\right)^{-1} \\ &\approx E_o\left(1 - \frac{E}{2mc^2}\right) \qquad \text{Iterate} \\ &\approx E_o\left[1 - \frac{E_o(1 - E/2mc^2)}{2mc^2}\right] \end{aligned}

[Alternate: Quadratic formula —

E^2 + \gamma E + \gamma E_o = 0 \quad \text{where } \gamma = 2mc^2

\Rightarrow\ E = \frac{\gamma}{2}\left[-1 \pm \left(1 + \frac{4E_o}{\gamma}\right)^{1/2}\right] \approx \frac{\gamma}{2}\left[-1 \pm \left(1 + \frac{2E_o}{\gamma} - \frac{2E_o^2}{\gamma^2}\right)\right]

since $E > 0$ take $+$

E = E_o(1 - \gamma)

Note: 2nd order needed in Binomial]

If $\delta \equiv \dfrac{E_o}{2mc^2} \ll 1$ then $\dfrac{EE_o}{(2mc^2)^2} \approx \delta^2$ is negligible

\Rightarrow\ \boxed{E \approx E_o\left(1 - \frac{E_o}{2mc^2}\right)}

(c) From part (b),

E_o - E = \frac{E_o^2}{2mc^2}

Using $E = hf$ , and writing $E_o^2 = hf_oE_o$ ,

h(f_o - f) = \frac{hf_oE_o}{2mc^2}

\frac{\Delta f}{f_o} = \frac{E_o}{2mc^2} = \frac{4\times10^{5}\,\mathrm{eV}}{2(2\times10^{11}\,\mathrm{eV})} = 10^{-6}

(d) For $v \ll c$ , Doppler shift says

\frac{v}{c} \approx \frac{\Delta\lambda}{\lambda} \qquad \text{where } c = \lambda f

Since $f_o > f$ , the observer must move towards the photon

\Delta\lambda = c\,\Delta\!\left(\frac{1}{f}\right) \approx -\frac{\Delta f}{f^2}\,c

[Alternate: $c = \lambda f$ ; $dc = 0 = \lambda\,df + f\,d\lambda$ ; $d\lambda = -\lambda\dfrac{df}{f}$ ; $\dfrac{d\lambda}{\lambda} = -\dfrac{df}{f}$ ; minus sign means that a decrease in $\lambda$ corresponds to increase in frequency!]

$\therefore$ (in magnitude)

\begin{aligned} \frac{v}{c} &\approx \frac{\Delta f}{f^2}\,\frac{c}{\lambda} \\ &= \frac{\Delta f}{f} \end{aligned}

\begin{aligned} v &= \frac{\Delta f}{f_o}\,c \\ &= 10^{-6}\,c \qquad \left(\text{using } \frac{\Delta f}{f_o} = 10^{-6} \text{ from part (c)}\right) \\ &= 300\,\mathrm{m/s} \end{aligned}

Hand sketch for (b): before decay a circle labelled M' at rest; after decay a circle labelled M moving left with momentum p and a photon dot moving right, labelled photon.

Hand sketch for (d): a photon dot moving right with speed c and a stick-figure observer moving left towards it with speed v.

Оригинал в Архива: Semifinal 1997.pdf · официални решения: Semifinal 1997 Solutions.pdf