4 - radius of nucleus & mass spectrometer

8/12/2019 4 - Radius of Nucleus & Mass Spectrometer

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Nuclear Physics


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Recap

The famous Geiger-Marsden Alpha scattering

experiment (under Rutherfords guidance)

In 1909, Geiger and Marsden were

studying how alpha particles are scattered

by a thin gold foil.

Alpha

source

Thin gold foil


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Geiger-Marsden

As expected, most alpha particles were

detected at very small scattering angles

Alpha particles

Thin gold foil Small-angle

scattering


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Geiger-Marsden

To their great surprise, they found that

some alpha particles (1 in 20 000) had

very large scattering angles

Alpha particles

Thin gold foil Small-angle

scattering

Large-angle

scattering


5/55

Explaining Geiger and Marsdens results

The results suggested that the positive (repulsive) charge must beconcentrated at the centre of the atom. Most alpha particles do not pass

close to this so pass undisturbed, only alpha particles passing very close to

this small nucleus get repelled backwards (the nucleus must also be very

massive for this to happen).

Remember on this scale, if the

nucleus is 2 cm wide, the atom

would be 200 m wide!


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Rutherford did the calculations!

Rutherford calculated theoretically the

number of alpha particles that should be

scattered at different angles (using

Coulombs law). He found agreement withthe experimental results if he assumed the

atomic nucleus was confined to a diameter

of about 10-15metres.


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Closest approach (new)

Using the idea of energy conservation, Rutherford found

it is possible to calculate the closest an alpha particle

can get to the nucleus during a head-on collision.

Alpha particle

nucleus


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Closest approach

Initially, the alpha particle has kinetic

energy = mu2

K.E. = mu2


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Closest approach

At the point of closest approach, the

particle reaches a distance rfrom the

nucleus and comes momentarily to rest.

r

K.E. = 0


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Closest approach

All the initial kinetic energyhas been

transformedto electrical potential energy.

K.E. = 0

r


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Closest approach

Using the formula for electrical potential energywhich is derived from Coulombs law

Kinetic energy lost = Electrical potential

mu2= 1 q1q2

4o r

K.E. = 0

r


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Closest approach

Rearranging we get;

r = 1 q1q2

4o mu2

In this case k is the Coulomb constant

K.E. = 0

r

umkr qq

2

21

21


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Closest approach

For an alpha particle, m = 6.7 x 10-27kg, q1= 2 x (1.6 x

10-19C) and u is around 2 x 107m.s-1. If the foil is made

of gold, q2is 79 x (1.6 x 10-19C).

r = 1 q1q2

4o mu2

r = 1 x (2 x 1.6 x 10-19C) x (79 x 1.6 x 10-19C)

4x 8.85 x 10-12Fm-1 x 6.7 x 10-27kg x (2 x 107m.s-1)2

r = 2.7 x 10-14m


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Rutherfords findings From the equation Rutherford found that the alpha particles

approached nuclei to within 3.2 x 10-14 m when the foil was made of

gold. Hence the radius of the gold nucleus must be less than this

value.

For silver atoms the closest distance of approach was found to be

2 x 10-14 m.

From these results, Rutherford concluded that the positive charge is

concentrated in a small sphere, that he called the nucleus. Whose

radii is no greater than about 10-14 m.

A new length was suggested for these small values called the

femtometer (fm) where 1 fm 10

-15

m. Since the time of Rutherfords scattering experiments, a multitude of

other experiments have shown that most nuclei are roughly

spherical and have an average radius given by;

A is the mass number and r0is a constant

equal to 1.2 x 10-15 m.

31

0Arr


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Video

Closest approach, and mass spectrometer
http://www.youtube.com/watch?v=PiTtRaqqJAshttp://www.youtube.com/watch?v=PiTtRaqqJAs


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The mass spectrometer

A VERY useful machine for measuring the

masses of atoms (ions) and their relative

abundances.


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The Mass Spectrometer

ions

produced

velocity selector

ionsaccelerated

Region of

magnetic field

detector

ion beam


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Electrons are produced at a

hot cathode and accelerated

by an electric field. Collisionwith gas particles produces

ions.

Substance to be tested

is turned into a gas byheating.


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Ions acclerated by

an electric fieldNarrow slit through

which ions pass


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Velocity selector In many experiments involving the motion of charged particles it is

important to have particles moving with the same velocity. This is

achieved by using a combination of an electric field and a magnetic

field, as shown below. The orientation is known as crossed fields.


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Mass Spec.Velocity selector

A uniform electric field vertically downward is provided by a pair ofcharged parallel plates, while a uniform magnetic field is applied

perpendicular to the page. For q positive, the magnetic force qvB is

upward and the electric force qE is downward.

If the fields are chosen such that the magnetic force balances the

electric force, the particle moves in a straight horizontal line throughthe region of the fields.


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Mass Spec.Velocity selector

If we adjust the size of the two fields so that qvB = qE, we get

Only those particles having this speed pass undeflected through the

perpendicular electric and magnetic fields. The magnetic force

acting on particles with speeds greater than this is stronger than the

electric force, and these particles are deflected upward. Thosehaving speeds less than this are deflected downward.

B

Ev


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Ions enter the velocity selectorwhich

contains an electric field (down the

page) and magnetic field (into the

page) at right anglesto each other.

By choosing a suitable value for the

magnetic field the ions continue in a

straight line. i.e. The force produced

by the electric filed (qE) is equal to

the force produced by the magnetic

field (Bqv). qE = Bqv


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qE = Bqv

orv = E/B

This means that only ions with a

specific velocity pass through this

region. (Hence velocity selector)


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The selected ions all with the

same velocity (but different

masses of course) enter the

second region of magnetic

field (also into the page). They

are deflected in a circular

path.


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Heavier ions continue forward

and hit the sides, as do ions

that are too light. Only ions of

one particluar mass reach thedetector.

The radius of the circle is

given by R = mv/qB so the

mass can be calculated from

m = RqB/v


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The magnetic field can be varied

so that ions of different mass can

be detected (higher B would

mean that ions of larger mass

could be directed at the detector).

In an experiment B is gradually

varied from low to high so that all

ions pass through and their

relative abundance measured.


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The detector can measure the

numbers of ions detected,hence giving an idea of

relative abundance of different

ions.


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The mass spectrometer is

particulary useful foridentifying isotopesof the

same element.


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Nuclear energy levels

We have seen previously that electrons exist inspecific energy levels around the atom.

There is evidence that energy levels existinside the nucleus too.


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When a nucleus decays by

emitting an alpha particle or

a gamma ray, the particles

or photons emitted are onlyat specific energies (there is

not a complete range of

energies emitted, only

certain specific levels).


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An alpha particle or photon thus has an

energy equal to the difference between

energy levels of the nucleus.

energy

levels in235U

(MeV)

51.57

0.051

0.013

0.000


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In the alpha decay of 239Pu to 235U, the plutonium

nucleus with an energy of 51.57 MeV can decay into

Uranium at 3 different energy levels.

energy

levels in235U

(MeV)

51.57

0.051

0.013

0.000

Plutonium-239


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If the 239Pu (51.57 MeV) decays to the ground stateof235U (0 MeV), an alpha particle of energy 51.57 MeV is

emitted.

energy

levels in235U

(MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Alpha emission (51.57

MeV)


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If the 239Pu (51.57 MeV) decays to the 2nd excited state

of 235U (0.051 MeV), an alpha particle of energy 51.57-

0.051 = 51.52 MeV is emitted. The uranium nucleus is

now in an excited state so can decay further by gamma

emission to the ground state.energy

levels in235U

(MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Alpha emission

(51.52 MeV)

Gamma

emission (0.051

MeV)


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In fact the nucleus could decay first to the 0.013 level,

and then the ground state, thus emitting two gamma

photons.

energy

levels in235U

(MeV)

51.57

0.051

0.013

0.000

Plutonium-239

Alpha emission

(51.52 MeV)

Gammaemission (0.038

MeV)

Gamma

emission

(0.013MeV)


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Notice that the energy of the emitted alpha

particles can only have certain discrete

values and that the gamma rays are

observed with a discrete spectra.

Remember gamma rays are part of the

EMS.

This is evidence that the nucleus like theatom is a quantum system and so has

discrete energy levels.


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Radioactive decay

Beta () and positron decay (+)

decay


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Radioactive decay

In beta decay, a neutron in the nucleus

decays into a proton, an electron and an

antineutrino.

n p + e + ve1

0

1

1

0

-1

0

0


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Radioactive decay

In positron decay, a proton in the nucleus

decays into a neutron, a positron (the

antiparticle of the electron) and a neutrino.

p n + e + ve

http://www.youtube.com/watch?v=lAAmAbJvvJg

1

1

1

0

0

+1

0

0
http://www.youtube.com/watch?v=lAAmAbJvvJghttp://www.youtube.com/watch?v=lAAmAbJvvJg


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The antineutrino

The antineutrinoin beta decaywas not

detected until 1953, although its presence

had been predicted theoretically.

n p + e + ve1

0

1

1

0

-1

0

0


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The antineutrino

The mass of the neutron is bigger than

that of the proton and electron together.

n p + e + ve

1.008665u(1.007276 + 0.0005486)u = 0.00084u

1

0

1

1

0

-1

0

0

1.008665 u 1.007276 u 0.0005486 u


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The antineutrino

This corresponds (using E = mc2) to an

energy of 0.783 MeV.

n p + e + ve1

0

1

1

0

-1

0

0


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The antineutrino

This extra energy should show up as kinetic energy ofthe products (proton and electron). Since the electronshould carry most of the kinetic energy away, so weshould observe electrons with an energy of about 0.783

MeV.

n p + e + ve

In fact we observe electrons with a rangeof energiesfrom zero up to 0.783 MeV.

1

01

1

0

-1

0

0


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The antineutrino

Where is the missing energy? In 1933 Wolfgang Pauli and

Enrico Fermi hypothesized the existence of a third very light

particle produced during the decay. Enrico Fermi coined the

term neutrino for the little neutral one

n p + e + ve1

0

1

1

0

-1

0

0


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Radioactive decay

The rate of decay (number of nuclei that will

decay per second) is proportional to the

number of undecayed nuclei

dN = -N

dt

= decay constant = the likelihood that an

individual particle will decay in a second


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The radioactive decay law

If the number of nuclei present in a sample at t =

0 is N0, the number Nstill present at time tlater

is given by

N = Noe-t

where is the decay constant (the probabilitythat a nucleus will decay in unit time)


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Half-life

After one half-life, the number of original

nuclei present is equal to N0/2. Putting this

into the radioactive decay law;

N0/2 = N0e(-t)where tis the half-life


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Activity

A =N = N = Noe-t

t


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Measuring half-life

For short half-lives, the half life can usually

be measured directly.


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Measuring half-life

For longer half lifes, values of activitycan

be measured and the decay law can be

used to calculate and thus t.

Measure the activity A and chemicallyfindthe number of atoms of the isotope.

UseA = N and then t= ln2

4 - radius of nucleus & mass spectrometer

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