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Chemistry Regents June 2010 - Question 01 PDF Print E-mail
Written by The Chemistry Wizard   

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Correct Answer: 1

Why: The gold foil experiment (also known as the Gelger-Marsden experiment or Rutherford experiment), was carried out by Hans Geiger and Ernest Marsden in 1909 to help determine the structure of the atom. The prevailing theory of atomic structure at the time was the plum pudding model, proposed by J. J. Thomson. This model proposed that atoms were composed of a positively charged cloud in which electrons, or negatively charged particles, were suspended. The gold foil experiment was designed to determine the distribution of electrons in the plum pudding model.

The gold foil experiment involved firing alpha particles through gold foil and onto a sheet Zinc sulfide, which would light up when struck by the alpha particles. Alpha particles are positively charged particles composed of two protons and two neutrons. When these particles were fired at the gold foil, according to the plum pudding model of the atom, the alpha particles would pass right through the gold foil with minor deflections occurring. It was assumed that there were no particles with enough mass or charge to deflect accelerated alpha particles at an angle greater than 90 degrees. However, what happened was a small number of the alpha particles were deflected at much greater angles than expected.

This led to two significant findings: (1.) rather than the distributed negative charge in a positive cloud, the positive charge carried most of the mass of an atom (to be able to deflect an alpha particle, it had to have high mass and be positively charged as like charges repel), and (2.) the nucleus was actually very small (or more collisions would have occurred).

With this knowledge the alpha particles would have mostly passed through the golf foil with a few being deflected at greater than 90 degree angles as they collided with the small, high mass and positively charged nuclei of the atoms.

Answering the Question:

To correctly answer this question, it is important to read the question carefully. The question states that the gold foil experiment led to the conclusion that each atom is composed mostly of empty space. If the atom were composed mostly of empty space, then particles fired at it would not be trapped, as suggested by answer (2). Particles becoming trapped would not suggest mostly empty space, but that atoms were densely packed with electrons. The positively charged alpha particles would have been attracted to the densely packed, negatively charged electrons and would become trapped.

Answer (3) would suggest a very large positively charged nuclei taking up most of the volume of the atom. The positive nuclei would then repel the positively charged alpha particles.

Answer (4) would not be possible as electrons are negatively charged and alpha particles are positively charged. The presence of opposite charges would result in strong forces of attraction and not, as the answer suggests, deflect the alpha particles.
Last Updated on Tuesday, 07 June 2011 16:36
 
Chemistry Regents June 2010 - Question 27 PDF Print E-mail
Written by The Chemistry Wizard   

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Answer: (3)

Why? There are many types of high-energy emissions known today, common types of which are alpha particles, beta particles and gamma rays.  An alpha particle (represented by the Greek letter α) is the equivalent of a He2+ ion. It consists of two protons and two neutrons, this gives it both charge and mass as the protons are positively charged and the nuclide of an atom contains most of its mass.

A beta particle (represented by the Greek letter β) is a high-speed electron or positron generated by radioactive decay of unstable nuclei. As both electrons and positrons have a charge and mass (although miniscule), a beta particle would not be the correct answer.

A Gamma ray (represented by the Greek letter γ) is high-energy electromagnetic radiation. Gamma rays are produced from processes such as fission (the main process of nuclear reactors) and fusion (the reaction that powers the sun) and are composed of high-energy waves with frequencies above 1019Hz. This type of radiation is very dangerous and is sometimes referred to as penetrating radiation, as it will penetrate materials that would prevent the passage of alpha and beta particles. However gamma rays do not have mass or charge and therefore, fits the requirements of the question.

A positron, which in its simplest definition is a positively charged electron or positively charged beta particle. Therefore, a positron does have charge and mass and would not meet the requirements of the question.

Answering the Question: If an understanding of the definition of each answer was unknown, answer (1) and answer (2) are both particulate, suggesting at the very least both have mass, eliminating both as correct answers. The only possible answers would be gamma rays and positrons, if neither was known a good guess would be gamma ray simply because it is a ray, suggesting no mass and possibly no charge.
 
Chemistry Regents June 2010 - Question 29 PDF Print E-mail
Written by The Chemistry Wizard   

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Answer: (4)

Why? A radioactive isotope is usually the result of a difference in the number of protons and neutrons, which causes instability within the nucleus of atoms, hence the term nuclear. When an element undergoes radioactive decay, it does so to achieve a more stable nuclide.

Elements that undergo alpha decay (α) lose two protons and two neutrons, or a 42He2+ ion plus energy. This decay helps to bring the neutron to proton ratio down and hence makes the nuclide more stable.

Beta decay occurs in one of two forms, β+ and β-. β- results from having a high ratio of neutrons to protons. It occurs when one of the neutrons in the nucleus splits to form a proton, which remains in the nucleus and an electron, which is emitted at high velocity. In effect for beta minus you lose a neutron gain a proton. A β+ emission occurs when the atom has a low neutron to proton ratio, in this decay a proton splits to form a neutron, which remains in the nucleus, and a positron, this form of decay is not as common as beta minus decay. In effect beta plus decay loses a proton and gains a neutron.

Gamma radiation (γ) is short wave high-energy electromagnetic radiation. A change in the nuclide of the atom does not occur with the release of gamma radiation. Gamma rays are high in energy and are produced from processes such as nuclear fusion and fission or when an electron and a positron come in contact with each other.

With this knowledge the answers can be analyzed to determine if any of the forms of decay would produce a more stable nuclide.

Answering the Question:

To answer this question, it is important to understand what it is that makes an atom radioactive. Answer (1) is the hydrogen isotope tritium (3H), which is an unstable isotope of hydrogen. However, as γ rays do not result in a decrease in the number of neutrons to create a more stable form of hydrogen, γ decay is unlikely.

31H → 31H + γ radiation

Answer (2) is potassium 42 and is said to undergo β+ decay, which involves the loss on a proton and the gaining of a neutron. This would result in Argon 42 being formed, which would probably be less stable that potassium 42.

4219K → 4218Ar + 0+1e (β+ particle)

Answer (3) is nitrogen 16; α decay would result in the loss of two neutrons and two protons, resulting in Boron 12, which would probably be unstable.

167N → 125B + 42He (α particle)

Answer (4) is phosphorous 32; β- decay would result in the loss of a neutron, and the gaining of a proton. This would result in the formation of sulfur 32, which is the naturally occurring form of sulfur, and would, therefore, have a stable nuclide.

3215P → 3216S + 0-1e (β- particle)
Last Updated on Tuesday, 07 June 2011 08:41
 
Chemistry Regents June 2010 - Question 50 PDF Print E-mail
Written by The Chemistry Wizard   

 

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Answer: (1)

Why? All known living organisms contain the element carbon. Carbon is considered the element of life and is most abundant in the form 12C. In this form there are an equivalent number of protons and neutrons, which make up the nucleus of the atom. Atoms however, sometimes have the same number of protons (used to define the element), but differing numbers of neutrons (used to define the nuclide). In the case of carbon, it is defined by having 6 protons and 6 neutrons, in the case of 12C. There are other nuclides of carbon, which although they occur naturally, are present in much smaller quantities; one such is 14C, which has 6 protons and 8 neutrons. Carbon 14 is slightly radioactive and has a half-life of approximately 5740±40 years. Carbon 14 is present in living organisms, starting in plants with a ratio to 12C similar to what is present in nature. This information along with the half life of carbon was exploited by Willard Libby et al in 1949 to determine the approximate age of dead organisms, a process termed carbon dating. Carbon dating is determined mathematically by determining the ratio of 14C to 12C in the dead organism. Radioactive dating methods are dependent on knowing the half-life of the materials used to carry out the dating. The half-life of the material is the time it takes for the material to decay and decrease by half.

Answering the Question:

To answer this question it helps to understand the role of carbon in living organisms. However, if that were not known the concept of carbon dating tends to be a familiar one.  Answer (1) suggests the use of C-14 and C-12 for dating remains of once living organisms. Once-living organisms would be composed mostly of carbon. This would make answer (1) with the 2 nuclides of carbon a good guess for use as a means for dating organisms.

Answer (2) suggests nuclides of cobalt be used for carbon dating. Cobalt is used in some living organisms, but not all and is quite harmful in large quantities. Co-60 is radioactive and releases gamma radiation, making it very harmful in all but the minutest quantities. Co-60 also has a half-life of approximately six years; this would not make it very suitable for dating.

Answer (3) suggests the use of iodine (I-131) and xenon (Xe-131), both of which are highly unlikely. Xenon in particular, is a noble gas and, therefore, although present, is very stable in terms of chemical bonding and half-life. This would not make it suitable for dating, as mathematically there would be very little basis on which to carry out dating calculations. This would not make answer (3) suitable for dating.

Answer (4) suggests the use of uranium (U-238) and lead (Pb-206) both of which are extremely harmful to living organisms and the environment in general. Although the half life of both elements is known they would be far too dangerous in living organisms to be present in any significant quantities and would not make them suitable for dating living organisms. Therefore, answer (1) is the best possible answer.
 
Chemistry Regents June 2010 - Question 62 PDF Print E-mail
Written by The Chemistry Wizard   

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Answer: Fission Reaction

Why? Equation 1 represents a fission reaction; this reaction type usually involves the splitting of a heavy atom into two smaller atoms and resulting neutrons. This reaction is very exothermic and is the type of reaction used to power nuclear power plants.

There are a number of heavy atoms, which are used, in nuclear fission; among them are Uranium, Plutonium and Thorium.

Answering the Question:

To answer the question, it is important to analyze Equation 1. This equation at first would appear odd when compared to a balanced chemical equation. However, when looked at more closely it is observed that the equation is still balanced. The mass of reactant is equal to the mass of the product.
 
Chemistry Regents June 2010 - Question 63 PDF Print E-mail
Written by The Chemistry Wizard   

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Answer: 23592U + 10n → 9238Sr + 14254Xe + 210n + energy

Answering the Question:

Equation 2 will behave like any other chemical equation. That is, the total mass of reactant will be equal to the total mass of product. In equation 2, uranium 235 undergoes nuclear fission to produce strontium 92, xenon 142, two neutrons and energy. Analyzing the mass number shows the mass number of the reactants is equal to the mass number of the products.

23592U + 10n → 9238Sr + 14254Xe + 210n + energy

235 + 1 → 92+142+(2x1), therefore

236 → 236, so not mass is lost.

The same holds true for the number of protons distributed throughout the reaction.

23592U + 10n → 9238Sr + 14254Xe + 210n + energy

92+0 → 38+54+0, therefore

92 → 92

So the number of protons also remains the same. To determine the possible products of nuclear fission, it is important to remember that both sides of the chemical reaction must be equal. This means that the mass number and total number of protons can be calculated from the given data.

The mass number would be calculated by:

23592U + 10n → 9238Sr + ______ + 210n + energy

235U + 1n → 92Sr+ unknown +(2x1) therefore

236 = 94 + unknown

236 -94 = unknown

Unknown = 142.

The same method can be used to calculate the number of protons (bottom number).

23592U + 10n → 9238Sr + _______ + 210n + energy

U + 0 → Sr + unknown + 0

92+ 0 → 38 + unknown + 0

92 → 38 + unknown

Therefore, unknown = 54.

This means the unknown has a proton count or 54 and a mass number of 142. The only element with 54 protons is Xenon (Xe). Therefore, based on calculations the resulting atom could be drawn

14254Xe.

 
Chemistry Regents June 2010 - Question 64 PDF Print E-mail
Written by The Chemistry Wizard   

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Answer: Number of half life periods (n) = t/T, fraction remaining = 1/2n

Using the fraction remaining = 6/96 = 0.0625

Therefore, 0.0625 = 1/ (2n)

2n = 1/0.0625 = 16

n = 4

Using the value of n in the equation, number of half life periods (n) = t/T, then

n = total time/half life

Therefore, 4 = 7.36/T

T = 7.36/4 = 1.84

The half-life of the reaction is 1.84 seconds

The half-life of radioactive decay is the time it takes for the substance to reach half its original mass.  That means a substance undergoing decay will follow the pattern:

image064-table

The decaying atom will have a reduction of its mass with each successive nuclear decay. However, it also means that the decay happens in a predictable pattern, and as such can be calculated when sufficient information is supplied.

 


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