topic 13: quantum and nuclear physics 13.2 nuclear physics

13.2.1 Explain how the radii of nuclei may be estimated from charges particle scattering experiments.

13.2.2 Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer.

13.2.3 Describe one piece of evidence for the existence of nuclear energy level.

Topic 13: Quantum and nuclear physics 13.2 Nuclear physics

Explain how the radii of nuclei may be estimated from charged particle scattering experiments.

As you may recall from Topic 7, Rutherford determined that the bulk of the atom’s positive charge (and mass) is located in a very small central nucleus have a radius of about 10-15 m.

If we analyze a head-on collision between an alpha particle and a nucleus, we can obtain a rough value for the diameter of a nucleus.

A nucleus has a charge of positive Ze and an alpha particle has a charge of positive 2e.

As the alpha particle approaches the nucleus it will be slowed down, stopped (for an instant), and reversed by the Coulomb force.


FYIReview Topic 7 for much of what is in this option.


The alpha particle feels a coulomb potential caused by the nucleus given by V = kQ/r or

V = kZe/r.

If the alpha particle approaches from infinity, the work needed to stop it at rmax is given by

W = qV = 2eV = 2kZe2/rmax.

From the work-kinetic energy theorem (W = ∆EK) we get W = EK-EK,0. Since at rmax, EK = 0, W = EK,0.

where EK,0 is the initial kinetic energy of the particle.


rmax

rmax = 2kZe2/EK,0 nuclear radius

EXAMPLE: Alpha particles having a kinetic energy of EK = 5 MeV bombard a gold nucleus and rebound straight back. Estimate the radius of the gold nucleus.

SOLUTION: Gold (Au) has Z = 79.

The alpha particle’s kinetic energy must be converted to joules:

EK,0 = (5106 eV)(1.610-19 J/eV) = 810-13 J.

Then rmax = 2kZe2/EK,0

= 2(9109)(79)(1.610-19)2/810-13

= 510-14 m.



rmax

rmax = 2kZe2/EK,0 nuclear radius

Why is the actual radius of the nucleus less than rmax.

Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer.

Recall the mass spectrometer in which an atom is stripped of its electrons and accelerated through a voltage into a magnetic field.

Scientists determined through the use of such a device that hydrogen nuclei came in three different masses:

Since the charge of the hydrogen nu-cleus is e, scientists postulated the existence of a neutral particle called the neutron.



While in the magnetic field the charged particle feels a centripetal force caused by the magnetic field of

Fc = qvB sin .But the angle between v and B is 90° so that

sin = 1.Since Fc = mv2/r then

mv2/r = qvB so that


r = mv/(qB) mass and radius



PRACTICE: Track X shows the deflection of a singly-charged carbon-12 ion in the deflection chamber of a mass spectrometer. Which path best shows the deflection of a singly-charged carbon-14 ion? Assume both ions travel at the same speed.SOLUTION: Since carbon-14 is heavier, it will have a bigger radius than carbon-12.Since its mass is NOT twice the mass of carbon-12, it will NOT have twice the radius.

B-field Source

A

B

C

D

X



EXAMPLE: A hydrogen ion (proton) is accelerated through a potential difference of V = 475 V and projected into a mass spectrometer having a magnetic field strength of 0.250 T.

(a) What is the velocity of the proton after its acceleration?

(b) What is its radius of curvature in the spec?

SOLUTION:

(a) EK = qV = (1.610-19)(475) = 7.610-17 J.

Then 7.610-17 = (1/2)mv2 = (1/2)(1.6710-27)v2 and

v = 3.0105 ms-1.

(b) r = mv/(qB)

= (1.6710-27)(3.0105)/[(1.610-19)(0.250)]

= 0.013 m (1.3 cm)

Describe one piece of evidence for the existence of nuclear energy levels.

Recall that electrons in an atom moving from an excited state to a de-excited state release a photon.

The emission spectra of de-exciting atoms show the existence of atomic energy levels.

In exactly the same way, de-exciting nuclei also release photons which also produce spectra - only with very high energy photons called gamma rays: 234Pu* 234Pu +


Atomic spectral lines.

Nuclear spectral lines.

Radioactive decay

13.2.4 Describe + decay, including the existence of the neutrino.

13.2.5 State that the radioactive decay law is an exponential function and define the decay constant.

13.2.6 Derive the relationship between the decay constant and the half-life.

13.2.7 Outline methods for measuring the half-life of an isotope.

13.2.8 Solve problems involving radioactive half-life.


Radioactive decay

Describe beta plus (+) decay including the existence of the neutrino.

There are two types of beta () particle decay:In - decay, a neutron becomes a proton and an electron is emitted from the nucleus. 14C 14N + + e-

In + decay, a proton becomes a neutron and a positron is emitted from the nucleus. 10C 10B + + e+

In short, a beta particle is either an electron or an anti-electron.


-

+

Radioactive decay

Describe beta plus (+) decay including the existence of the neutrino.

In contrast to the alpha particle, it was discovered that beta particles could have a large variety of kinetic energies.

In order to conserve energy it was postulated that another particle called a neutrino was created to carry the additional EK needed to balance the energy.


Medium

Medium

Slow

Fast

Same total energy

Radioactive decay

State the radioactive decay law as an exponential function and define the decay constant.

The higher the initial population of a radioactive material, the more decays there will be in a time interval.

But each decay decreases the population.

Hence the decay rate decreases over time for a fixed sample and it is an exponential decrease.

where N0 is the initial population, N is the new one, t is the time, and is the decay constant.


Time in half-livesRadioactive material

remaining

N = Noe-t population decay

Radioactive decay

Derive the relationship between decay constant and half-life.


EXAMPLE: Show that the relationship between half-life and decay constant is given by T1/2 = ln 2/.SOLUTION: Use N = Noe-t. Then N = N0/2 when t = T1/2.

Then N = Noe-t

N0/2 = Noe-T

(1/2) = e-T

ln(1/2) = -T1/2

-ln(1/2) = T1/2

ln 2 = T1/2

Exponential decay function.

Substitution.Cancel N0.

ln x and ex are inverses.

Multiply by -1.

-ln (1/x) = +ln x.

T1/2 = ln 2/ decay constant and half-life

Radioactive decay

Solve problems involving radioactive half-life.


EXAMPLE: The half-life of U-238 is 4.51010 y and for I-123 is 13.3 h. Find the decay constant for each radioactive nuclide.

SOLUTION:

Use T1/2 = ln 2/. Then = ln 2/T1/2.

For U-238 we have = ln 2/T1/2 = 0.693/4.51010 y = 1.510-11 y-1.

For I-123 we have = ln 2/T1/2 = 0.693/13.3 h = 0.052 h-1.

FYIThe decay constant is the probability of decay of a nucleus per unit time.

Radioactive decay

Outline methods for measuring the half-life of an isotope.

Rather than measuring the amount of remaining radioactive nuclide there is in a sample (which is hard to do) we measure instead the decay rate (which is much easier).

Decay rates are measured using various devices, most commonly the Geiger-Mueller counter.

Decay rates are measured in Becquerels (Bq).


1 Becquerel

1 Bq = 1 decay / second

Radioactive decay

Outline methods for measuring the half-life of an isotope.

The decay rate A is given by

The ∆N is the change in the number of nuclei, and is negative (the radioactive sample loses population with each decay).

The negative sign is in A = -∆N/∆t to make the activity A positive.

A = N shows that the activity is proportional to the remaining population of radioactive nuclei, which we learned in Topic 7.

Since N = Noe-t the last equation A = Noe-t is true.


A = -∆N/∆t = N = Noe-t decay rate or activity

PRACTICE:

Radioactive decay



Remember that the mass of the material does not change appreciatively during radioactive decay.

Nuclei are just transmuted.

PRACTICE:

Radioactive decay



If you look at the lower numbers you see that we are short a positive charge on the right:The only two particles with a positive charge (that we have studied) are the beta+ and the proton.Looking at the nucleon number we see that it must be the proton.

0e1

1p1

It is a proton.

PRACTICE:

Radioactive decay



The CO2 in the atmosphere has a specific percentage of carbon-14.The moment the wood dies, the carbon-14 is NOT replenished.Since the carbon-14 is always disintegrating and is NOT being replenished in the dead wood, its activity will decrease over time.

PRACTICE:

Radioactive decay



From Thalf = ln 2/ we get = ln 2/Thalf or = 0.693/5500 = 0.00013 y-1.

From A = N we see that in the beginning 9.6 = N0 and now 2.1 = N.

Thus N = N0e-t becomes 2.1 = 9.6e-t so that2.1/9.6 = e-t.

ln(2.1/9.6) = ln(e-t)

-1.5198 = -t

t = 1.5198/0.00013 = 12000 y

PRACTICE:

Radioactive decay



The activity would be too small to be reliable.

For this sample

A = 9.1e-t becomes

A = 9.1e-0.00013(20000) = 0.68 decay min-1.

topic 13: quantum and nuclear physics 13.2 nuclear physics

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