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IPhO 1989, Теория и експеримент — Задача 3. Problem 3

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Problems of the 20th International Physics Olympiad (Warsaw, 1989) · 16 юли 1989 г.

Abstract

The article contains problems given at the 20th International Physics Olympiad (1989) and their solutions. The 20th IPhO was the third IPhO organized in Warsaw, Poland.

¹ This article has been sent for publication in Physics Competitions in October 2003

² e-mail: gorzk@ifpan.edu.pl

Logo

The emblem of the XX International Physics Olympiad contains a picture that is a historical record of the first hypernuclear event observed and interpreted in Warsaw by M. Danysz and J. Pniewski. The collision of a high-energy particle with a heavy nucleus was registered in nuclear emulsion. Tracks of the secondary particles emitted in the event, seen in the picture (upper star), consist of tracks due to fast pions (“thin tracks”) and to much slower fragments of the target nucleus (“black tracks”). The “black track” connecting the upper star (greater) with the lower star (smaller) in the figure is due to a hypernuclear fragment, in this case due to a part of the primary nucleus containing an unstable hyperon Λ instead of a nucleon. Hyperfragments (hypernuclei) are a new kind of matter in which the nuclei contain not only protons and neutrons but also some other heavy particles.

In the event observed above the hyperon Λ, bound with nucleon, decays like a free particle through a week (slow) process only. This fact strongly suggested the existence of a new quantum number that could explain suppression of the decay, even in presence of nucleons. Indeed, this was one of the observations that, 30 months later, led to the concept of strangeness.

M. Danysz and J. Pniewski, Bull. Acad. Polon. Sci., 3(1) 42 (1952) and Phil. Mag., 44, 348 (1953). Later the same physicists, Danysz and Pniewski, discovered the first case of a nucleus with two hyperons (double hyperfragment).

Увод

Theoretical problems (including solutions and marking schemes) were prepared especially for the 20th IPhO by Waldemar Gorzkowski. The experimental problem (including the solution and marking scheme) was prepared especially for this Olympiad by Andrzej Kotlicki. The problems were refereed independently (and many times) by at least two persons after any change was made in the text to avoid unexpected difficulties at the competition. This work was done by:

First Problem:
Andrzej Szadkowski, Andrzej Szymacha, Włodzimierz Ungier

Second Problem:
Andrzej Szadkowski, Andrzej Szymacha, Włodzimierz Ungier, Stanisław Woronowicz

Third Problem:
Andrzej Rajca, Andrzej Szymacha, Włodzimierz Ungier

Experimental Problem:
Krzysztof Korona, Anna Lipniacka, Jerzy Łusakowski, Bruno Sikora

Several English versions of the texts of the problems were given to the English-speaking students. As far as I know it happened for the first time (at present it is typical). The original English version was accepted (as a version for the students) by the leaders of the Australian delegation only. The other English-speaking delegations translated the English originals into English used in their countries. The net result was that there were at least four English versions. Of course, physics contained in them was exactly the same, while wording and spelling were somewhat different (the difference, however, were not too great).

This article is based on the materials quoted at the end of the article and on personal notes of the author.

Условие

The problem concerns investigation of transforming the electron microscope with magnetic guiding of the electron beam (which is accelerated with the potential difference $U$ = 511 kV) into a proton microscope (in which the proton beam is accelerated with the potential difference $-U$ ). For this purpose, solve the following two problems:

A. An electron after leaving a device, which accelerated it with the potential difference $U$ , falls into a region with an inhomogeneous field $\boldsymbol{B}$ generated with a system of stationary coils $L_1$ , $L_2$ , … , $L_n$ . The known currents in the coils are $i_1$ , $i_2$ , … , $i_n$ , respectively.

What should the currents $i_1'$ , $i_2'$ , … , $i_n'$ in the coils $L_1$ , $L_2$ , … , $L_n$ be, in order to guide the proton (initially accelerated with the potential difference $-U$ ) along the same trajectory (and in the same direction) as that of the electron?

HINT: The problem can be solved by finding a condition under which the equation describing the trajectory is the same in both cases. It may be helpful to use the relation:

\boldsymbol{p}\frac{d}{dt}\boldsymbol{p} = \frac{1}{2}\frac{d}{dt}\boldsymbol{p}^2 = \frac{1}{2}\frac{d}{dt}p^2.

B. How many times would the resolving power of the above microscope increase or decrease if the electron beam were replaced with the proton beam? Assume that the resolving power of the microscope (i.e. the smallest distance between two point objects whose circular images can be just separated) depends only on the wave properties of the particles.

Assume that the velocities of the electrons and protons before their acceleration are zero, and that there is no interaction between own magnetic moment of either electrons or protons and the magnetic field. Assume also that the electromagnetic radiation emitted by the moving particles can be neglected.

NOTE: Very often physicists use 1 electron-volt (1 eV), and its derivatives such as 1 keV or 1 MeV, as a unit of energy. 1 electron-volt is the energy gained by the electron that passed the potential difference equal to 1 V.

Perform the calculations assuming the following data:

Rest energy of electron: $E_e = m_ec^2 = 511\ \mathrm{keV}$
Rest energy of proton: $E_p = m_pc^2 = 938\ \mathrm{MeV}$

Отговори

Решение

PART A

At the beginning one should notice that the kinetic energy of the electron accelerated with the potential difference $U$ = 511 kV equals to its rest energy $E_0$ . Therefore, at least in the case of the electron, the laws of the classical physics cannot be applied. It is necessary to use relativistic laws.

The relativistic equation of motion of a particle with the charge $e$ in the magnetic field $\mathbf{B}$ has the following form:

\frac{d}{dt}\mathbf{p} = \mathbf{F}_L

where $\mathbf{p} = m_0\gamma\mathbf{v}$ denotes the momentum of the particle (vector) and

\mathbf{F}_L = e\mathbf{v} \times \mathbf{B}

is the Lorentz force (its value is $evB$ and its direction is determined with the right hand rule). $m_0$ denotes the (rest) mass of the particle and $v$ denotes the velocity of the particle. The quantity $\gamma$ is given by the formula:

\gamma = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}

The Lorentz force $\mathbf{F}_L$ is perpendicular to the velocity $\mathbf{v}$ of the particle and to its momentum $\mathbf{p} = m_0\gamma\mathbf{v}$ . Hence,

\mathbf{F}_L \cdot \mathbf{v} = \mathbf{F}_L \cdot \mathbf{p} = 0.

Multiplying the equation of motion by $\mathbf{p}$ and making use of the hint given in the text of the problem, we get:

\frac{1}{2}\frac{d}{dt}p^2 = 0.

It means that the value of the particle momentum (and the value of the velocity) is constant during the motion:

p = m_0v\gamma = \mathrm{const}; \qquad v = \mathrm{const}.

The same result can be obtained without any formulae in the following way:

The Lorentz force $\mathbf{F}_L$ is perpendicular to the velocity $\mathbf{v}$ (and to the momentum $p$ as $\mathbf{p} = m_0\gamma\mathbf{v}$ ) and, as a consequence, to the trajectory of the particle. Therefore, there is no force that could change the component of the momentum tangent to the trajectory. Thus, this component, whose value is equal to the length of $\mathbf{p}$ , should be constant: $\mathbf{p} = \mathrm{const}$ . (The same refers to the component of the velocity tangent to the trajectory as $\mathbf{p} = m_0\gamma\mathbf{v}$ ).

Let $s$ denotes the path passed by the particle along the trajectory. From the definition of the velocity, we have:

\frac{ds}{dt} = v.

Using this formula, we can rewrite the equation of motion as follows:

v\frac{d}{ds}\mathbf{p} = \frac{ds}{dt}\frac{d}{ds}\mathbf{p} = \frac{d}{dt}\mathbf{p} = \mathbf{F}_L,

\frac{d}{ds}\mathbf{p} = \frac{\mathbf{F}_L}{v}.

Dividing this equation by $p$ and making use of the fact that $p$ = const, we obtain:

v\frac{d}{ds}\frac{\mathbf{p}}{p} = \frac{\mathbf{F}_L}{vp}

and hence

\frac{d}{ds}\mathbf{t} = \frac{\mathbf{F}_L}{vp}

where $\mathbf{t} = \mathbf{p}/p = \mathbf{v}/v$ is the versor (unit vector) tangent to the trajectory. The above equation is exactly the same for both electrons and protons if and only if the vector quantity:

\frac{\mathbf{F}_L}{vp}

is the same in both cases.

Denoting corresponding quantities for protons with the same symbols as for the electrons, but with primes, one gets that the condition, under which both electrons and protons can move along the same trajectory, is equivalent to the equality:

\frac{\mathbf{F}_L}{vp} = \frac{\mathbf{F}'_L}{v'p'}.

However, the Lorentz force is proportional to the value of the velocity of the particle, and the directions of any two vectors of the following three: $\mathbf{t}$ (or $\mathbf{v}$ ), $\mathbf{F}_L$ , $\mathbf{B}$ determine the direction of the third of them (right hand rule). Therefore, the above condition can be written in the following form:

\frac{e\mathbf{B}}{p} = \frac{e'\mathbf{B}'}{p'}.

Hence,

\mathbf{B}' = \frac{e}{e'}\frac{p'}{p}\mathbf{B} = \frac{p'}{p}\mathbf{B}.

This means that at any point the direction of the field $\mathbf{B}$ should be conserved, its orientation should be changed into the opposite one, and the value of the field should be multiplied by the same factor $p'/p$ . The magnetic field $\mathbf{B}$ is a vector sum of the magnetic fields of the coils that are arbitrarily distributed in the space. Therefore, each of this fields should be scaled with the same factor $-p'/p$ . However, the magnetic field of any coil is proportional to the current flowing in it. This means that the required scaling of the fields can only be achieved by the scaling of all the currents with the same factor $-p'/p$ :

i'_n = -\frac{p'}{p}i_n.

Now we shall determine the ratio $p'/p$ . The kinetic energies of the particles in both cases are the same; they are equal to $E_k = e\left|U\right| = 511$ keV. The general relativistic relation between the total energy $E$ of the particle with the rest energy $E_0$ and its momentum $p$ has the following form:

E^2 = E_0^2 + p^2c^2

where c denotes the velocity of light.

The total energy of considered particles is equal to the sum of their rest and kinetic energies:

E = E_0 + E_k.

Using these formulae and knowing that in our case $E_k = e\left|U\right| = E_e$ , we determine the momenta of the electrons ( $p$ ) and the protons ( $p’$ ). We get:

a) electrons:

(E_e + E_e)^2 = E_e^2 + p^2c^2,

p = \frac{E_e}{c}\sqrt{3}.

b) protons

(E_p + E_e)^2 = E_p^2 + p'^2c^2,

p' = \frac{E_e}{c}\sqrt{\left(\frac{E_p}{E_e} + 1\right)^2 - \left(\frac{E_p}{E_e}\right)^2}.

Hence,

\frac{p'}{p} = \frac{1}{\sqrt{3}}\sqrt{\left(\frac{E_p}{E} + 1\right)^2 - \left(\frac{E_p}{E_e}\right)^2} \approx 35.0

and

i'_n = -35.0\, i_n.

It is worthwhile to notice that our protons are 'almost classical', because their kinetic energy $E_k (= E_e)$ is small compared to the proton rest energy $E_p$ . Thus, one can expect that the momentum of the proton can be determined, with a good accuracy, from the classical considerations. We have:

E_e = E_k = \frac{p'^2}{2m_p} = \frac{p'^2c^2}{2m_pc^2} = \frac{p'^2c^2}{2E_p},

p' = \frac{1}{c}\sqrt{2E_eE_p}.

On the other hand, the momentum of the proton determined from the relativistic formulae can be written in a simpler form since $E_p/E_e \gg 1$ . We get:

p' = \frac{E_e}{c}\sqrt{\left(\frac{E_p}{E_e} + 1\right)^2 - \left(\frac{E_p}{E_e}\right)^2} = \frac{E_e}{c}\sqrt{2\frac{E_p}{E_e} + 1} \approx \frac{E_e}{c}\sqrt{2\frac{E_p}{E_e}} = \frac{1}{c}\sqrt{2E_eE_p}.

In accordance with our expectations, we have obtained the same result as above.

PART B

The resolving power of the microscope (in the meaning mentioned in the text of the problem) is proportional to the wavelength, in our case to the length of the de Broglie wave:

\lambda = \frac{h}{p}

where $h$ denotes the Planck constant and $p$ is the momentum of the particle. We see that $\lambda$ is inversely proportional to the momentum of the particle. Therefore, after replacing the electron beam with the proton beam the resolving power will be changed by the factor $p/p' \approx 1/35$ . It means that our proton microscope would allow observation of the objects about 35 times smaller than the electron microscope.

Marking scheme

the relativistic equation of motion — 1 point
independence of $p$ and $v$ of the time — 1 point
identity of $e\mathbf{B}/p$ in both cases — 2 points
scaling of the fields and the currents with the same factor — 1 point
determination of the momenta (relativistically) — 1 point
the ratio of the momenta (numerically) — 1 point
proportionality of the resolving power to $\lambda$ — 1 point
inverse proportionality of $\lambda$ to $p$ — 1 point
scaling of the resolving power — 1 point

Remarks and typical mistakes in the pupils' solutions

Some of the participants tried to solve the problem by using laws of classical mechanics only. Of course, this approach was entirely wrong. Some students tried to find the required condition by equating "accelerations" of particles in both cases. They understood the "acceleration" of the particle as a ratio of the force acting on the particle to the "relativistic" mass of the particle. This approach is incorrect. First, in relativistic physics the relationship between force and acceleration is more complicated. It deals with not one "relativistic" mass, but with two "relativistic" masses: transverse and longitudinal. Secondly, identity of trajectories need not require equality of accelerations.

The actual condition, i.e. the identity of $e\mathbf{B}/p$ in both cases, can be obtained from the following two requirements:

1° in any given point of the trajectory the curvature should be the same in both cases;

2° in the vicinity of any given point the plane containing a small arc of the trajectory should be oriented in space in both cases in the same way.

Most of the students followed the approach described just above. Unfortunately, many forgot about the second requirement (they neglected the vector character of the quantity $e\mathbf{B}/p$ ).

Acknowledgement

I would like to thank very warmly to Prof. Jan Mostowski and Dr. Andrzej Wysmołek for reading the text of this article and for valuable critical remarks. I express special thanks to Dr. Andrzej Kotlicki for critical reviewing the experimental part of the article and for a number of very important improvements.

Literature

[1] Waldemar Gorzkowski and Andrzej Kotlicki, XX Międzynarodowa Olimpiada Fizyczna - cz. I, Fizyka w Szkole nr 1/90, pp. 34 - 39

[2] Waldemar Gorzkowski, XX Międzynarodowa Olimpiada Fizyczna - cz. II, Fizyka w Szkole nr 2/3-90, pp. 23 - 32

[3] XX International Physics Olympiad - Proceedings of the XX International Physics Olympiad, Warsaw (Poland), July 16 - 24, 1989, ed. by W. Gorzkowski, World Scientific Publishing Company, Singapore 1990 [ISBN 981-02-0084-6]

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