Question 1

The half-life of americium-241 is 457 years.What is the activity in curies of a 0.010 g sample of Am-241? (1 Ci = 3.700 × 10¹⁰ atom/s)

Accepted Answer

The activity (A) of a radioactive sample is given by the formula A = (λN)/3.7x10^10, where λ is the decay constant and N is the number of atoms. The decay constant λ can be calculated from the half-life (T_1/2) using the formula λ = ln(2)/T_1/2. Given the half-life of americium-241 is 457 years, and the mass of the sample is 0.010 g, we can calculate the number of atoms (N) using Avogadro's number and the molar mass of Am-241. After calculating N and λ, we find the activity and convert it to curies (Ci), leading to the correct answer of approximately 0.032 Ci.

Question 2

All of the following statements are true EXCEPT

Accepted Answer

The statement "nuclear fission evolves more energy than nuclear fusion" is false. Nuclear fusion actually releases more energy than nuclear fission.

Question 3

The following reaction is an example of ____. 
${ }_{92}^{236} \mathrm{U} \rightarrow{ }_{39}^{96} \mathrm{Y}+{ }_{53}^{136} \mathrm{I}+4{ }_{0}^{1} n$

Accepted Answer

The reaction shown involves the splitting of a large uranium nucleus into smaller nuclei (yttrium and iodine) along with the release of neutrons, which is characteristic of a fission reaction.

Question 4

The energy change for a nuclear reaction can be calculated from which of the following equations?

Accepted Answer

The correct equation for calculating the energy change for a nuclear reaction is given by $\Delta E = c^2\Delta m$, where $\Delta E$ is the change in energy, $c$ is the speed of light in a vacuum, and $\Delta m$ is the change in mass. This corresponds to choice D, which seems to be a typographical representation of the famous equation $E=mc^2$ by Albert Einstein, indicating the relationship between mass (m) and energy (E) with $c$ being the speed of light.

Question 5

An extremely sensitive method has recently been developed for measuring $ { }_{6}^{14} \mathrm{C} $ concentrations in organic materials.What is this method?

Accepted Answer

The measurement of $ { }_{6}^{14} \mathrm{C} $ concentrations in organic materials requires the use of mass spectrometry. This is because $ { }_{6}^{14} \mathrm{C} $ is a radioactive isotope and its concentration in materials is extremely low. Mass spectrometry is a sensitive analytical technique that can accurately measure the concentration of this isotope in organic materials.

Question 6

The half-life of $ { }_{17}^{36} \mathrm{Cl} $ is 9.5 × 10¹² s.How many atoms of $ { }_{17}^{36} \mathrm{Cl} $ are present in a sample with an activity of 3.9 × 10⁹ atoms/s?

Accepted Answer

The answer of The half-life of  \( { }_{17}^{36}...

Question 7

A 1.00 g sample of carbon taken from a living plant has an activity of 13.6 atoms/min.If 1.00 g of an organic sample is found to have an activity of 6.27 atoms/min,what is the age of the sample? The decay constant of $ { }_{6}^{14} \mathrm{C} $ is 1.21 × 10−⁴ yr−¹.

Accepted Answer

The answer of A 1.00 g sample of carbon taken...

Question 8

A 1.00 g sample of carbon taken from a fossilized plant has an activity of 0.19 atoms/s.What mass of $ { }_{6}^{14} \mathrm{C} $ is present? The decay constant of ${ }_{6}^{14} \mathrm{C}$ is 3.8 × 10?¹² s?¹.

Accepted Answer

The answer of A 1.00 g sample of carbon taken...

Question 9

The point of maximum stability in the binding energy curve occurs in the vicinity of which one of the following isotopes?

Accepted Answer

The point of maximum stability in the binding energy curve represents the most tightly bound nucleus, and it occurs at the element with the highest binding energy per nucleon. This is because nuclei with higher binding energies per nucleon have lower potential energy and are therefore more stable. ${ }_{26}^{56}\mathrm{Fe}$ is the isotope with the highest binding energy per nucleon, making it the most stable.

Question 10

The following nuclear reaction is an example of ____. 
$4_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2_{1}^{0} e$

Accepted Answer

This reaction involves the fusion of four hydrogen atoms to form one helium atom, making it an example of fusion.

Question 11

Carbon dating can be applied to organic materials up to 5.0 × 10⁴ years old.What percentage of the original $ { }_{6}^{14} \mathrm{C} $ is present in a sample after this much time? The half-life of $ { }_{6}^{14} \mathrm{C} $ is 5730 yr.

Accepted Answer

Using the formula for exponential decay, we can calculate the fraction of $ { }_{6}^{14} \mathrm{C} $ remaining after 5.0 × 10⁴ years as: $N/N_{0} = (1/2)^{(t/T)}}$ where: - N is the remaining amount of $ { }_{6}^{14} \mathrm{C} $ - N0 is the initial amount of $ { }_{6}^{14} \mathrm{C} $ - t is the time elapsed (5.0 × 10⁴ years in this case) - T is the half-life of $ { }_{6}^{14} \mathrm{C} $ (5730 years) $N/N_{0} = (1/2)^{(t/T)}} = (1/2)^{(5.0\times10^{14}\mathrm{years}/5730\mathrm{years})} \approx 0.0024$ This means that approximately 0.24% of the original $ { }_{6}^{14} \mathrm{C} $ is present in a sample after 5.0 × 10⁴ years. Therefore, the answer is B.

Question 12

The half-life of mercury-203 is 46.6 days.What percentage of mercury-203 remains in a sample after 365 days?

Accepted Answer

The percentage of mercury-203 remaining after 365 days can be calculated using the formula for radioactive decay, which is: remaining amount = initial amount * (1/2)^(time/half-life). Here, time = 365 days and half-life = 46.6 days. Plugging in the values: remaining amount = 100% * (1/2)^(365/46.6) ≈ 0.439%.

Question 13

How long does it take for 105 mg of tritium, $ { }_{1}^{3} \mathrm{H} $ ,to decay to 35.0 mg? The half-life of tritium is 12.3 yr.

Accepted Answer

The answer of How long does it take for 105...

Question 14

How many atoms decay in 12 hours in a 3.5 × 10−⁸ Ci source? (1 Ci = 3.700 × 10¹⁰ atom/s)

Accepted Answer

Firstly, we need to convert the radioactive source activity from Ci to atoms per second:
3.5 × 10^18 Ci = 3.5 × 10^18 × 3.700 × 10^10 atoms/s = 1.295 × 10^29 atoms/s

Then, we can use the radioactive decay law to determine the number of atoms that decay in 12 hours:
N(t) = N(0) × e^(-λt), where N(t) is the number of atoms at time t, N(0) is the initial number of atoms, λ is the decay constant, and t is time.

Since the decay constant is related to the activity A as λ = A/N(0), we can rewrite the decay law as:
N(t) = N(0) × e^(-At/N(0))

Plugging in the values we have:
N(12 hours) = N(0) × e^(-At/N(0)) = (1.295 × 10^29 atoms/s) × e^[-(3.700 × 10^10 atoms/s) × (12 hours × 3600 s/h) / (6.022 × 10^23 atoms/mol)] = 5.540 × 10^17 atoms

Therefore, the number of atoms that decay in 12 hours is:
N(0) - N(12 hours) = (3.5 × 10^18 atoms) - (5.540 × 10^17 atoms) = 4.946 × 10^18 atoms, which is approximately equal to 5.6 × 10^17 atoms or 5.6 × 10^17. Thus, the closest option would be D) 5.6 × 10^17 atoms.

Question 15

The rate constant for the decay of copper-62 is 9.8 minutes−¹.What is the half-life of this isotope?

Accepted Answer

The half-life of an isotope is the time it takes for half of the atoms in a sample to decay. The rate constant for the decay of an isotope is the probability that an atom will decay per unit time. The half-life of an isotope is inversely proportional to the rate constant. Therefore, the half-life of copper-62 is 1/9.8 = 0.10 minutes.

Question 16

What is the relationship between the rate constant and the half-life for nuclear decay?

Accepted Answer

The relationship between the rate constant (k) and the half-life (t1/2) for nuclear decay is given by the equation k = ln(2)/t1/2. This equation is derived from the exponential decay equation A = A0e-kt, where A is the amount of radioactive material left after time t, and A0 is the initial amount. The half-life is the time it takes for half of the initial amount to decay, which can be solved for by setting A = 1/2A0 and t = t1/2. Substituting these values into the exponential decay equation and solving for k yields k = ln(1/2)/t1/2 = ln(2)/t1/2.

Question 17

The decay constant,k,for $ { }_{88}^{226} \mathrm{Ra} $ is 1.37 × 10−¹¹ s−¹.What mass of $ { }_{88}^{226} \mathrm{Ra} $ has an activity of 1.06 × 10⁹ atoms/s?

Accepted Answer

The answer of The decay constant,k,for  \( { }_{88}^{226}...

Question 18

Uranium-235 has a half-life of 7.04 × 10⁸ years.How many years will it take for 99.9% of a U-235 sample to decay?

Accepted Answer

The answer of Uranium-235 has a half-life of 7.04 ×...

Question 19

A 1.00 g sample of $ { }_{32}^{66} \mathrm{Ge} $ decays to 0.87 g in 1.9 hr.What is the half-life of $ { }_{32}^{66} \mathrm{Ge} $?

Accepted Answer

- First, we need to calculate how much of the sample has decayed using the following formula:

Amount of substance left = Initial amount of substance x (1/2)^(t/half-life)

Where t is the time elapsed and half-life is the time taken for half of the sample to decay.

- Let's plug in the given values:

0.87 g = 1.00 g x (1/2)^(1.9/half-life)

- Now, we can solve for the half-life:

(1/2)^(1.9/half-life) = 0.87/1.00

(1/2)^(1.9/half-life) = 0.87

Take the natural logarithm of both sides:

ln(1/2)^(1.9/half-life) = ln(0.87)

(1.9/half-life)ln(1/2) = ln(0.87)

(1.9/half-life) = ln(0.87)/ln(1/2)

half-life = 4.5 hr

Therefore, the correct answer is C.

Question 20

All of the following statements concerning radiocarbon dating are true EXCEPT

Accepted Answer

Radiocarbon dating is not used to determine the age of meteorites. It is used primarily to determine the age of organic materials on Earth.

Quiz 18: Nuclear Reactions

The half-life of americium-241 is 457 years.What is the activity in curies of a 0.010 g sample of Am-241? (1 Ci = 3.700 × 10¹⁰ atom/s)

All of the following statements are true EXCEPT

The following reaction is an example of ____. ${ }_{92}^{236} \mathrm{U} \rightarrow{ }_{39}^{96} \mathrm{Y}+{ }_{53}^{136} \mathrm{I}+4{ }_{0}^{1} n$

The energy change for a nuclear reaction can be calculated from which of the following equations?

An extremely sensitive method has recently been developed for measuring ${ }_{6}^{14} \mathrm{C}$ concentrations in organic materials.What is this method?

The half-life of ${ }_{17}^{36} \mathrm{Cl}$ is 9.5 × 10¹² s.How many atoms of ${ }_{17}^{36} \mathrm{Cl}$ are present in a sample with an activity of 3.9 × 10⁹ atoms/s?

A 1.00 g sample of carbon taken from a living plant has an activity of 13.6 atoms/min.If 1.00 g of an organic sample is found to have an activity of 6.27 atoms/min,what is the age of the sample? The decay constant of ${ }_{6}^{14} \mathrm{C}$ is 1.21 × 10−⁴ yr−¹.

A 1.00 g sample of carbon taken from a fossilized plant has an activity of 0.19 atoms/s.What mass of ${ }_{6}^{14} \mathrm{C}$ is present? The decay constant of ${ }_{6}^{14} \mathrm{C}$ is 3.8 × 10?¹² s?¹.

The point of maximum stability in the binding energy curve occurs in the vicinity of which one of the following isotopes?

The following nuclear reaction is an example of ____. $4_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2_{1}^{0} e$

Carbon dating can be applied to organic materials up to 5.0 × 10⁴ years old.What percentage of the original ${ }_{6}^{14} \mathrm{C}$ is present in a sample after this much time? The half-life of ${ }_{6}^{14} \mathrm{C}$ is 5730 yr.

The half-life of mercury-203 is 46.6 days.What percentage of mercury-203 remains in a sample after 365 days?

How long does it take for 105 mg of tritium, ${ }_{1}^{3} \mathrm{H}$ ,to decay to 35.0 mg? The half-life of tritium is 12.3 yr.

How many atoms decay in 12 hours in a 3.5 × 10−⁸ Ci source? (1 Ci = 3.700 × 10¹⁰ atom/s)

The rate constant for the decay of copper-62 is 9.8 minutes−¹.What is the half-life of this isotope?

What is the relationship between the rate constant and the half-life for nuclear decay?

The decay constant,k,for ${ }_{88}^{226} \mathrm{Ra}$ is 1.37 × 10−¹¹ s−¹.What mass of ${ }_{88}^{226} \mathrm{Ra}$ has an activity of 1.06 × 10⁹ atoms/s?

Uranium-235 has a half-life of 7.04 × 10⁸ years.How many years will it take for 99.9% of a U-235 sample to decay?

A 1.00 g sample of ${ }_{32}^{66} \mathrm{Ge}$ decays to 0.87 g in 1.9 hr.What is the half-life of ${ }_{32}^{66} \mathrm{Ge}$ ?

All of the following statements concerning radiocarbon dating are true EXCEPT

Quiz 18: Nuclear Reactions

The half-life of americium-241 is 457 years.What is the activity in curies of a 0.010 g sample of Am-241? (1 Ci = 3.700 × 1010 atom/s)

All of the following statements are true EXCEPT

The following reaction is an example of ____. 92236U→3996Y+53136I+401n{ }_{92}^{236} \mathrm{U} \rightarrow{ }_{39}^{96} \mathrm{Y}+{ }_{53}^{136} \mathrm{I}+4{ }_{0}^{1} n92236​U→3996​Y+53136​I+401​n

The energy change for a nuclear reaction can be calculated from which of the following equations?

An extremely sensitive method has recently been developed for measuring 614C { }_{6}^{14} \mathrm{C} 614​C concentrations in organic materials.What is this method?

The half-life of 1736Cl { }_{17}^{36} \mathrm{Cl} 1736​Cl is 9.5 × 1012 s.How many atoms of 1736Cl { }_{17}^{36} \mathrm{Cl} 1736​Cl are present in a sample with an activity of 3.9 × 109 atoms/s?

A 1.00 g sample of carbon taken from a living plant has an activity of 13.6 atoms/min.If 1.00 g of an organic sample is found to have an activity of 6.27 atoms/min,what is the age of the sample? The decay constant of 614C { }_{6}^{14} \mathrm{C} 614​C is 1.21 × 10−4 yr−1.

A 1.00 g sample of carbon taken from a fossilized plant has an activity of 0.19 atoms/s.What mass of 614C { }_{6}^{14} \mathrm{C} 614​C is present? The decay constant of 614C{ }_{6}^{14} \mathrm{C}614​C is 3.8 × 10?12 s?1.

The point of maximum stability in the binding energy curve occurs in the vicinity of which one of the following isotopes?

The following nuclear reaction is an example of ____. 411H→24He+210e4_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2_{1}^{0} e411​H→24​He+210​e

Carbon dating can be applied to organic materials up to 5.0 × 104 years old.What percentage of the original 614C { }_{6}^{14} \mathrm{C} 614​C is present in a sample after this much time? The half-life of 614C { }_{6}^{14} \mathrm{C} 614​C is 5730 yr.

The half-life of mercury-203 is 46.6 days.What percentage of mercury-203 remains in a sample after 365 days?

How long does it take for 105 mg of tritium, 13H { }_{1}^{3} \mathrm{H} 13​H ,to decay to 35.0 mg? The half-life of tritium is 12.3 yr.

How many atoms decay in 12 hours in a 3.5 × 10−8 Ci source? (1 Ci = 3.700 × 1010 atom/s)

The rate constant for the decay of copper-62 is 9.8 minutes−1.What is the half-life of this isotope?

What is the relationship between the rate constant and the half-life for nuclear decay?

The decay constant,k,for 88226Ra { }_{88}^{226} \mathrm{Ra} 88226​Ra is 1.37 × 10−11 s−1.What mass of 88226Ra { }_{88}^{226} \mathrm{Ra} 88226​Ra has an activity of 1.06 × 109 atoms/s?

Uranium-235 has a half-life of 7.04 × 108 years.How many years will it take for 99.9% of a U-235 sample to decay?

A 1.00 g sample of 3266Ge { }_{32}^{66} \mathrm{Ge} 3266​Ge decays to 0.87 g in 1.9 hr.What is the half-life of 3266Ge { }_{32}^{66} \mathrm{Ge} 3266​Ge ?

All of the following statements concerning radiocarbon dating are true EXCEPT

The half-life of americium-241 is 457 years.What is the activity in curies of a 0.010 g sample of Am-241? (1 Ci = 3.700 × 10¹⁰ atom/s)

The following reaction is an example of ____. ${ }_{92}^{236} \mathrm{U} \rightarrow{ }_{39}^{96} \mathrm{Y}+{ }_{53}^{136} \mathrm{I}+4{ }_{0}^{1} n$

An extremely sensitive method has recently been developed for measuring ${ }_{6}^{14} \mathrm{C}$ concentrations in organic materials.What is this method?

The half-life of ${ }_{17}^{36} \mathrm{Cl}$ is 9.5 × 10¹² s.How many atoms of ${ }_{17}^{36} \mathrm{Cl}$ are present in a sample with an activity of 3.9 × 10⁹ atoms/s?

A 1.00 g sample of carbon taken from a living plant has an activity of 13.6 atoms/min.If 1.00 g of an organic sample is found to have an activity of 6.27 atoms/min,what is the age of the sample? The decay constant of ${ }_{6}^{14} \mathrm{C}$ is 1.21 × 10−⁴ yr−¹.

A 1.00 g sample of carbon taken from a fossilized plant has an activity of 0.19 atoms/s.What mass of ${ }_{6}^{14} \mathrm{C}$ is present? The decay constant of ${ }_{6}^{14} \mathrm{C}$ is 3.8 × 10?¹² s?¹.

The following nuclear reaction is an example of ____. $4_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2_{1}^{0} e$

Carbon dating can be applied to organic materials up to 5.0 × 10⁴ years old.What percentage of the original ${ }_{6}^{14} \mathrm{C}$ is present in a sample after this much time? The half-life of ${ }_{6}^{14} \mathrm{C}$ is 5730 yr.

How long does it take for 105 mg of tritium, ${ }_{1}^{3} \mathrm{H}$ ,to decay to 35.0 mg? The half-life of tritium is 12.3 yr.

How many atoms decay in 12 hours in a 3.5 × 10−⁸ Ci source? (1 Ci = 3.700 × 10¹⁰ atom/s)

The rate constant for the decay of copper-62 is 9.8 minutes−¹.What is the half-life of this isotope?

The decay constant,k,for ${ }_{88}^{226} \mathrm{Ra}$ is 1.37 × 10−¹¹ s−¹.What mass of ${ }_{88}^{226} \mathrm{Ra}$ has an activity of 1.06 × 10⁹ atoms/s?

Uranium-235 has a half-life of 7.04 × 10⁸ years.How many years will it take for 99.9% of a U-235 sample to decay?

A 1.00 g sample of ${ }_{32}^{66} \mathrm{Ge}$ decays to 0.87 g in 1.9 hr.What is the half-life of ${ }_{32}^{66} \mathrm{Ge}$ ?