Chp 20: Nuclear Flashcards

1
Q

Observations and conclusions of the alpha scattering experiment

A
  1. Charge distribution
    Experimentally, most of the alpha particles pass thru un-deviated. Some were deflected by a very small angle. Very few deviated by large angle, cuz got scattered by a strong positive charge, which means strong positive charge got concentrated in a very small space within the atom. so, atom has large amount of empty space except at the centre
  2. Mass distribution
    Very few alpha particles undergo a large deflection and even lesser alpha particles undergo a 180-degree deflection. By COM, since got particles undergo a 180-degree deflection implies mass of the gold nucleus must be very large than that of alpha particles.
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2
Q

Explain the phenomena: In the alpha scattering experiment, only a small proportion of the alpha particles are deflected through angles greater than 90 degrees.

A

In atom, most vol is empty space where mass is concentrated in nucleus. Nucleus vry small than atom size

Hence when alpha particles are incident on the atoms of the metal foil, most of them pass through the atom with minimal scattering.

Only a few alpha particles pass vry close to or are incident on the nucleus of the atom, are scattered with vry large angles due to electrostatic repulsion cuz mass of the nucleus much larger than the alpha particle.

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3
Q

What is meant by binding energy of a nucleus and how it is related to the mass defect?

A

The binding energy of a nucleus: energy required to separate the nucleus into its constituent neutrons and protons such that they are separated to infinity.

The binding energy = mass defect x (speed of light)² in vacuum.

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4
Q

State the experimental evidence to suggest that the process of radioactive decay is:

Random:

Spontaneous:

A

Random: Fluctuations in the activity (not a smooth curve)

Spontaneous: rate of decay is not affected by external forces like temperature and pressure nor by any external stimuli.

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5
Q

Properties of Alpha, Beta and Gamma Radiations

A
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6
Q

Explain the phenomena: A radioactive source emits high energy beta particles inside an aluminum container. Although the beta particles are absorbed, X-ray radiation is emitted from the container walls.

A
  1. Beta particles bombard Al and exp range of deceleration
  2. When charged particles exp acceleration or deceleration, they emit EMR
  3. Range of emitted radiation is within X-ray energies
  4. KE of beta particles sufficient to knock out inner shell electrons of lead, resulting in emission of characteristic x-ray wavelengths
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7
Q

Why although count rates are low, careful shielding is necessary?

A

Careful shielding of the source: prevent prolonged close-proximity exposure to the ionizing radiation (alpha,beta,gamma,x-ray..) that may damage DNA and cause biological harm to the body in long term.

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8
Q

Parent nucleus before radioactive decay is stationary, explain why after the decay, the product nucleus moves.

A

Since the parent nucleus is initially stationary, initial momentum = 0. By PCOM, since got no external force, after decay, total momentum must = 0. Hence, product nuclei must be moving with the same magnitude of momentum but opp direction as the alpha particle/beta particle/gamma ray emitted.

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9
Q

What is 1 similarity and 1 difference between radioactive decay and nuclear fission.

A

Similarity: Both release energy.

Difference: Radioactive decay occurs spontaneously but nuclear fission need an external trigger, usually a neutron to split a heavy nucleus into 2 approximately sized daughter nuclei, releasing more neutrons.

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10
Q

Why is energy released per nucleon during nuclear fusion more than nuclear fission?

A

Looking at the BE/nucleon graph, the upward slope is steeper than the downward slope. Therefore, the increase in BE/nucleon for fusion is more than the increase in BE/nucleon for fission. Hence, …

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11
Q

What quantities must be conserved in any nuclear process?

A

Total proton number, total nucleon number, total momentum, total mass-energy.

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12
Q

Explain why the fusion of nuclei with high nucleon numbers is not associated with a release of energy.

A

In downward sloping region of the BE/nucleon graph, when a fusion process occur, pdts have a smaller BE/nucleon compared to the reactants. Pdts LESS STABLE than rxts. Since total BE of pdts < total BE of the rxts, energy must be absorbed for this fusion reaction to occur

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13
Q

Explain why a release of energy occurs when there is an increase in the binding energy.

A

An increase in BE means BE/nucleon of pdt nuclei > rxt nucleus. Means that energy released when forming the products > energy required when breaking the reactants into constituent particles. As a result, there is a net energy released.

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14
Q

The two decay patterns of Bi-214 into Ti and Po each give rise to gamma ray photons. (gamma decay) Suggest why each of these photons has different energies.

A

γ ray photon is emitted when excited nucleus of Bi-214 disintegrates into a nucleus with a lower energy state.

Energy of the γ ray photon = diff in energy levels within the nucleus when transition takes place.

Ti and Po have diff nuclear structures, so got diff nuclear energy levels. Hence … are also different.

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15
Q

Explain how the consideration of the kinetic energy spectrum of beta particles during a beta decay provides evidence for the prediction of the existence of the neutrino/antineutrino.

A

From E= mc2 , the total energy released and total momentum per decay is constant.

If only daughter nuclei and the beta particles are emitted, ALL beta particles should be emitted with same energy.

However, since got range of kinetic energies of beta particles, there must be a 3rd particle emitted known as the neutrino/antineutrino to possess the remaining energy and momentum

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16
Q

Explain why measuring the mass and then the activity of a sample to find the decay constant only works well for nuclides with small decay constants.

A

small decay constant = long half-life.

During the measurement process, the activity of the nuclide is relatively constant, can assume that the mass and activity is measured at the same instant.

17
Q

Explain why when determining half-life, it is better to use the part of the graph where the curve is steep?

A

At the steeper part of the graph, the activity of the sample has smaller percentage uncertainty.

So calculating half-life of the isotope will have smaller percentage uncertainty