Съдържание

IPhO 1988, Теория и експеримент — Задача 3. THEORY 3 Recombination of Positive and Negative Ions in Ionized Gas

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Всички материали от IPhO 1988

19th International Physics Olympiad - 1988 Bad Ischl / Austria

Условие

Introduction

A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit).

Let this ion be represented by the symbol $A^{(Z-1)+}$

Constants:

electric field constant $\varepsilon_O = 8{,}85\cdot 10^{-12}\ \frac{\mathrm{A\cdot s}}{\mathrm{V\cdot m}}$

elementary charge $e = \pm 1{,}602\cdot 10^{-19}\ \mathrm{A\cdot s}$

$q^2 = \frac{e^2}{4\cdot\pi\cdot\varepsilon_O} = 2{,}037\cdot 10^{-28}\ \mathrm{J\cdot m}$

Planck´s constant $\hbar = 1{,}054\cdot 10^{-34}\ \mathrm{J\cdot s}$

(rest) mass of an electron $m_e = 9{,}108\cdot 10^{-31}\ \mathrm{kg}$

Bohr´s atomic radius $r_B = \frac{\hbar}{m\cdot q^2} = 5{,}92\cdot 10^{-11}\ \mathrm{m}$

Rydberg´s energy $E_R = \frac{q^2}{2\cdot r_B} = 2{,}180\cdot 10^{-18}\ \mathrm{J}$

(rest) mass of a proton $m_P\cdot c^2 = 1{,}503\cdot 10^{-10}\ \mathrm{J}$

3.1 Assume that the ion which has just one electron left the shell.

$A^{(Z-1)+}$ is in the ground state.

In the lowest energy state, the square of the average distance of the electron from the nucleus or $r^2$ with components along x-, y- and z-axis being $(\Delta x)^2$ , $(\Delta y)^2$ and $(\Delta z)^2$ respectively and $r_O^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ and also the square of the average momentum by $p_O^2 = (\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2$ , whereas $\Delta p_x \ge \frac{\hbar}{2\cdot\Delta x}$ , $\Delta p_y \ge \frac{\hbar}{2\cdot\Delta y}$ and $\Delta p_z \ge \frac{\hbar}{2\cdot\Delta z}$ .

Write inequality involving $(p_O)^2\cdot(r_O)^2$ in a complete form.

3.2 The ion represented by $A^{(Z-1)+}$ may capture an additional electron and consequently emits a photon.

Write down an equation which is to be used for calculation the frequency of an emitted photon.

3.3 Calculate the energy of the ion $A^{(Z-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles:

3.3.A The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$ .

(ie. $\frac{1}{r_O}$ ; $r_0$ is given in the problem).

3.3.B In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $(p_O)^2\cdot(r_O)^2 \approx (\hbar)^2$

3.4 Calculate the energy of the ion $A^{(Z-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $A^{(Z-1)+}$ . Given the average distance of each of the two electrons in the outermost shell (same as $r_0$ given in 3.3) denoted by $r_1$ and $r_2$ , assume the average distance between the two electrons is given by $r_1+r_2$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie.

$p_1^2\cdot r_1^2 \approx \hbar^2$ and $p_2^2\cdot r_2^2 \approx \hbar^2$

hint: Make use of the information that in the ground state $r_1 = r_2$

3.5 Consider in particular the ion $A^{(Z-2)+}$ is at rest in the ground state when capturing an additional electron and the captured electron is also at rest prior to the capturing. Determine the numerical value of Z, if the frequency of the emitted photon accompanying electron capturing is $2{,}057\ .\ 10^{17}$ rad/s. Identify the element which gives rise to the ion.

Отговори

Решение

3.1

$r_0^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$

$p_0^2 = (\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2$

since

$\Delta p_x \ge \frac{\hbar}{2\cdot\Delta x}$ $\qquad$ $\Delta p_y \ge \frac{\hbar}{2\cdot\Delta y}$ $\qquad$ $\Delta p_z \ge \frac{\hbar}{2\cdot\Delta z}$

gives

p_0^2 \ge \frac{\hbar^2}{4}\cdot\left[\frac{1}{(\Delta x)^2} + \frac{1}{(\Delta y)^2} + \frac{1}{(\Delta z)^2}\right]

and

(\Delta x)^2 = (\Delta y)^2 = (\Delta z)^2 = \frac{r_0^2}{3}

thus $\boxed{p_0^2\cdot r_0^2 \ge \frac{9}{4}\cdot\hbar^2}$

3.2

$|\vec{v}_e|$ ...... speed of the external electron before the capture

$|\vec{V}_i|$ ....... speed of $A^{(Z-1)+}$ before capturing

$|\vec{V}_f|$ ...... speed of $A^{(Z-1)+}$ after capturing

$E_n = h.\nu$ ...... energy of the emitted photon

conservation of energy:

\frac{1}{2}\cdot m_e\cdot v_e^2 + \frac{1}{2}\cdot(M+m_e)\cdot V_i^2 + E\left[A^{(Z-1)+}\right] = \frac{1}{2}\cdot(M+2\cdot m_e)\cdot V_f^2 + E\left[A^{(Z-2)+}\right]

where $E[A^{(Z-1)+})$ and $E[A^{(Z-2)+}]$ denotes the energy of the electron in the outermost shell of ions $A^{(Z-1)+}$ and $A^{(Z-2)+}$ respectively.

conservation of momentum:

m_e\cdot\vec{v}_e + (M+m)\cdot\vec{V}_i = (M+2\cdot m_e)\cdot\vec{V}_f + \frac{h\cdot\nu}{c}\cdot\vec{1}

where $\vec{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.

3.3

Determination of the energy of $A^{(Z-1)+}$ :

potential energy $= -\frac{Z\cdot e^2}{4\cdot\pi\cdot\varepsilon_0\cdot r_0} = -\frac{Z\cdot q^2}{r_0}$

kinetic energy $= \frac{p^2}{2\cdot m}$

If the motion of the electrons is confined within the x-y-plane, principles of uncertainty in 3.1 can be written as

$r_0^2 = (\Delta x)^2 + (\Delta y)^2$

$p_0^2 = (\Delta p_x)^2 + (\Delta p_y)^2$

p_0^2 = \frac{\hbar^2}{4}\cdot\left[\frac{1}{(\Delta x)^2} + \frac{1}{(\Delta y)^2}\right] = \frac{\hbar^2}{4}\cdot\left[\frac{2}{r_0^2} + \frac{2}{r_0^2}\right] = \frac{\hbar^2}{4}\cdot\frac{4}{r_0^2}

thus

$p_0^2\cdot r_0^2 = \hbar^2$

E\left[A^{(Z-1)+}\right] = \frac{p_0^2}{2\cdot m_e} - \frac{Z\cdot q^2}{r_0} = \frac{\hbar^2}{2\cdot m_e\cdot r_e} - \frac{Z\cdot q^2}{r_0}

Energy minimum exists, when $\frac{dE}{dr_0} = 0$ .

Hence

\frac{dE}{dr_0} = -\frac{\hbar^2}{m_e\cdot r_e^3} + \frac{Z\cdot q^2}{r_0^2} = 0

this gives $\frac{1}{r_0} = \frac{Z\cdot q^2\cdot m_e}{\hbar^2}$

hence

E\left[A^{(Z-1)+}\right] = \frac{\hbar^2}{2\cdot m_e}\cdot\left(\frac{Z\cdot q^2\cdot m_e}{\hbar}\right)^2 - Z\cdot q^2\cdot\frac{Z\cdot q^2\cdot m_e}{\hbar^2} = -\frac{m_e}{2}\cdot\left(\frac{Z\cdot q^2}{\hbar}\right)^2 = -\frac{q^2\cdot Z^2}{2\cdot r_B} = -E_R\cdot Z^2

\boxed{E\left[A^{(Z-1)+}\right] = -E_R\cdot Z^2}

3.4

In the case of $A^{(Z-1)+}$ ion captures a second electron

potential energy of both electrons $= -2\cdot\frac{Z\cdot q^2}{r_0}$

kinetic energy of the two electrons $= 2\cdot\frac{p^2}{2\cdot m} = \frac{\hbar^2}{m_e\cdot r_0^2}$

potential energy due to interaction between the two electrons $= \frac{q^2}{\left|\vec{r}_1 - \vec{r}_2\right|} = \frac{q^2}{2\cdot r_0}$

E\left[A^{(Z-2)+}\right] = \frac{\hbar^2}{m_e\cdot r_0^2} - \frac{2\cdot Z\cdot q^2}{r_0^2} + \frac{q^2}{2\cdot r_0}

total energy is lowest when $\frac{dE}{dr_0} = 0$

hence

0 = -\frac{2\cdot\hbar^2}{m_e\cdot r_0^3} + \frac{2\cdot Z\cdot q^2}{r_0^3} - \frac{q^2}{2\cdot r_0^2}

hence

\frac{1}{r_0} = \frac{q^2\cdot m_e}{2\cdot\hbar^2}\cdot\left(2\cdot Z - \frac{1}{2}\right) = \frac{1}{r_B}\cdot\left(Z - \frac{1}{4}\right)

E\left[A^{(Z-2)+}\right] = \frac{\hbar^2}{m_e}\cdot\left(\frac{q^2\cdot m_e}{2\cdot\hbar^2}\right)^2 - \frac{q^2\cdot\left(2\cdot Z - \frac{1}{2}\right)}{\hbar}\cdot\frac{q^2\cdot m_e\cdot\left(2\cdot Z - \frac{1}{2}\right)}{2\cdot\hbar}

E\left[A^{(Z-2)+}\right] = -\frac{m_e}{4}\cdot\left[\frac{q^2\cdot\left(2\cdot Z - \frac{1}{2}\right)}{\hbar}\right]^2 = -\frac{m_e\cdot\left[q^2\cdot\left(Z - \frac{1}{4}\right)\right]^2}{\hbar^2} = -\frac{q^2\cdot\left(Z - \frac{1}{4}\right)^2}{\hbar^2}

this gives

\boxed{E\left[A^{(Z-2)+}\right] = -2\cdot E_R\cdot\left(Z - \frac{1}{4}\right)^2}

3.5

The ion $A^{(Z-1)+}$ is at rest when it captures the second electron also at rest before capturing.

From the information provided in the problem, the frequency of the photon emitted is given by

\nu = \frac{\omega}{2\cdot\pi} = \frac{2{,}057\cdot 10^{17}}{2\cdot\pi}\ \mathrm{Hz}

The energy equation can be simplified to $E\left[A^{(Z-1)+}\right] - E\left[A^{(Z-2)+}\right] = \hbar\cdot\omega = h\cdot\nu$

that is

-E_R\cdot Z^2 - \left[-2\cdot E_R\cdot\left(Z - \frac{1}{4}\right)^2\right] = \hbar\cdot\omega

putting in known numbers follows

2{,}180\cdot 10^{-18}\cdot\left[-Z^2 + 2\cdot\left(Z - \frac{1}{4}\right)^2\right] = 1{,}05\cdot 10^{-34}\cdot 2{,}607\cdot 10^{17}

this gives

$Z^2 - Z - 12{,}7 = 0$

with the physical sensuous result $Z = \frac{1+\sqrt{1+51}}{2} = 4{,}1$

This implies Z = 4, and that means Beryllium

Оригинал в Архива: ipho88.pdf