Question 1

Charcoal found under a stone at Stonehenge, England, has a carbon-14 activity that is 0.60 that of new wood. How old is the charcoal? (The half-life of carbon-14 is 5730 years.)

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Question 2

The radiochemist, Will I. Glow, studied thorium-232 and found that 2.82 × 10^-7 moles emitted 8.42 × 10⁶ α particles in one year. What is the decay constant for thorium-232?

Accepted Answer

The decay constant, denoted by λ, is the fraction of atoms that decay per unit time. It is related to the number of decays per unit time, denoted by N, through the equation N = λN0, where N0 is the initial number of atoms.

In this problem, we are given the number of decays per unit time (8.42 × 10^16 α particles per year) and the number of atoms (2.82 × 10^17 moles of thorium-232). We need to find the decay constant.

First, we need to convert the number of moles to the number of atoms. One mole of an element contains Avogadro's number of atoms, which is approximately 6.022 × 10^23. Therefore, 2.82 × 10^17 moles of thorium-232 contain:

2.82 × 10^17 moles × 6.022 × 10^23 atoms/mole = 1.698 × 10^41 atoms

Using the equation N = λN0, we can solve for λ:

N = λN0
8.42 × 10^16 α particles/year = λ (1.698 × 10^41 atoms)
λ = 8.42 × 10^16 α particles/year / 1.698 × 10^41 atoms
λ = 4.96 × 10^-11 year^-1

Therefore, the best choice is B: 4.96 × 10^-11 year^-1.

Question 3

A mass of 6.02 × 10²⁶ amu is equivalent to

Accepted Answer

1 amu is approximately 1.66053906660 × 10^-27 kg, so 6.02 × 10^26 amu is approximately 1 kg.

Question 4

Polonium-208 is an alpha emitter with a half-life of 2.90 years. How many milligrams of polonium from an original sample of 2.00 mg will remain after 8.00 years?

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Question 5

What is the nuclear binding energy per nucleon for ? (1 kg = 6.022 × 10²⁶ amu; C = 2.99792458 × 10⁸ m/s) Particle Mass (amu) 24.985839 1.007276 1.008665

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Question 6

Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation. (1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; C = 2.99792458 × 10⁸ m/s)

Accepted Answer

We start by writing the balanced equation for the decay of polonium-214:

214Po → 210Pb + 4He

The atomic masses of polonium-214, lead-210, and helium-4 are 214.0 amu, 210.0 amu, and 4.0026 amu, respectively. Therefore, the mass of one mole of polonium-214 is 214.0 g/mol, and the mass of one polonium-214 nucleus is 214.0/6.022×1023 g.

The mass difference between the reactant (polonium-214) and the products (lead-210 and helium-4) is:

Δm = (214.0 - 210.0 - 4.0026) g/mol = -0.0026 g/mol

Using Einstein's famous equation, E = mc2, we can calculate the energy released as a result of this mass difference:

E = Δm × c2

where c is the speed of light. Substituting the values, we get:

E = (-0.0026 g/mol) × (2.99792458 × 108 m/s)2 = -2.466 × 10-11 J/mol

The negative sign indicates that energy is released in the decay process.

Therefore, the correct answer is C: 8.76 × 1011 J/mol (rounded to two significant figures).

Question 7

Iodine-131, t_1/2= 8.0 days, is used in the diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-131 initially emits 9.95 × 10¹⁸ β particles per day, how long will it take for the activity to drop to 6.22 × 10¹⁷ β particles per day?

Accepted Answer

We can use the radioactive decay law to solve for the time it takes for the activity to drop to a certain level:

A(t) = A0*(1/2)^(t/t1/2)

where A(t) is the activity at time t, A0 is the initial activity, t1/2 is the half-life, and t is the time elapsed.

We can rearrange this equation to solve for t:

t = t1/2*(log(A(t)/A0)/log(1/2))

Plugging in the given values, we get:

t = 8.0*(log(6.22E17/9.95E18)/log(1/2)) = 31.9 days

Rounding to the nearest whole number, the answer is C) 32 days.

Question 8

What is the nuclear binding energy per nucleon of uranium-234? [1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; c = 2.99792458 × 10⁸ m/s]

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Question 9

The nuclear binding energy per nucleon is defined by which of the following?

Accepted Answer

The nuclear binding energy per nucleon is defined as the energy released when a nucleus is formed from its individual nucleons. This energy is then divided by the total number of nucleons in the nucleus to obtain the binding energy per nucleon. The correct formula for this is given by option A as it calculates the binding energy per nucleon directly. Option B and D are incorrect because they do not accurately represent the definition of nuclear binding energy per nucleon. Option C is also incorrect because it does not provide a way to calculate the binding energy per nucleon.

Question 10

An isotope with a high value of N/Z will tend to decay through

Accepted Answer

Isotopes with high N/Z ratios are neutron-rich and beta decay involves the conversion of a neutron into a proton, which decreases the N/Z ratio. Therefore, isotopes with high N/Z ratios tend to undergo beta decay to increase the proton to neutron ratio. Isotopes with high N/Z ratios are neutron-rich and unstable, so they tend to decay through beta (β) decay, in which a neutron is converted to a proton and an electron (β particle) is emitted. This leads to a decrease in the N/Z ratio and a stabilization of the nucleus. α decay typically occurs in heavy, proton-rich isotopes, while positron decay and electron capture are more common in isotopes with low N/Z ratios. γ decay involves the emission of gamma radiation, but is not a type of radioactive decay.

Question 11

What fraction of radioactive atoms remains in a sample after six half-lives?

Accepted Answer

Each half-life cuts the amount of remaining radioactive atoms in half. After six half-lives, the fraction of radioactive atoms remaining is (1/2)^6 which simplifies to 1/64. Therefore, the answer is E) 1/64. After one half-life, half of the radioactive atoms decay and are no longer present, leaving 1/2 of the original amount. After two half-lives, another half of the remaining radioactive atoms decay, leaving 1/2 of 1/2, or 1/4 of the original amount. Following this pattern, after six half-lives, the fraction remaining is 1/2 to the power of 6, which simplifies to 1/64. Therefore, the answer is E.

Question 12

Cesium-134 is a β emitter with a half-life of 2.0 years. How much of a 2.50-g sample of cesium-134 will remain after 10 years?

Accepted Answer

After 2 years (one half-life), half of the original cesium-134 will decay, leaving 1.25 g. After 4 years (two half-lives), only a quarter of the original cesium-134 will remain, leaving 0.625 g. Continuing in this manner, we can see that after 10 years (five half-lives), only 1/32 of the original cesium-134 will remain.

To calculate the final amount remaining, we can use the equation:

final amount = initial amount x (1/2)^(number of half-lives)

Plugging in the given values, we get:

final amount = 2.50 g x (1/2)^(5) = 0.078 g

Therefore, the answer is B) 0.078 g.

Question 13

What is the energy equivalent of 1 amu? (c = 2.99792458 × 10⁸ m/s)

Accepted Answer

The energy equivalent of 1 atomic mass unit (amu) is calculated using Einstein's equation E=mc². 1 amu is approximately 1.66053906660 × 10^-27 kg. Using c = 2.99792458 × 10^8 m/s, the energy is approximately 1.5 × 10^-10 J.

Question 14

What is the name for the difference between the mass of an atom and the sum of the masses of its constituent nucleons?

Accepted Answer

The term for the difference between the mass of an atom and the sum of the masses of its constituent nucleons is called the mass defect. It is caused by the release of energy during the formation of the nucleus, which is in accordance with Einstein's famous equation E=mc². The energy released contributes to the binding energy that holds the nucleus together, and the mass of the nucleus is hence less than the sum of its constituent particles. This is known as the mass-energy equivalence.

Question 15

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half life for ²³⁸U → ²⁰⁶Pb is 4.51 × 10⁹ yr.)

Accepted Answer

First, we need to find the ratio of Pb-206 to U-238 in the rock:
0.37 mg Pb-206 / 206 g/mol = 1.79 x 10^-6 mol Pb-206
0.95 mg U-238 / 238 g/mol = 3.99 x 10^-6 mol U-238

Now, we can use the ratio of U-238 to the total number of atoms in the rock to find the number of U-238 atoms:
3.99 x 10^-6 mol U-238 / (1 mol/6.022 x 10^23 atoms) = 2.4 x 10^12 atoms

However, we need to take the half-life of U-238 into account. We can use the formula for radioactive decay:
N = N0 (1/2)^(t/T)
where N is the final number of atoms, N0 is the initial number of atoms, t is the time elapsed, and T is the half-life.

We know that t is billions of years, but we don't need to use the exact number. Instead, we can use the fact that after one half-life, the number of U-238 atoms will be halved. We can keep halving the number until we get to the initial number of atoms (N0) that we're looking for:

N = 2.4 x 10^12 atoms / 2 = 1.2 x 10^12 atoms (after 1 half-life)
N = 1.2 x 10^12 atoms / 2 = 6.0 x 10^11 atoms (after 2 half-lives)
N = 6.0 x 10^11 atoms / 2 = 3.0 x 10^11 atoms (after 3 half-lives)
N = 3.0 x 10^11 atoms / 2 = 1.5 x 10^11 atoms (after 4 half-lives)
N = 1.5 x 10^11 atoms / 2 = 7.5 x 10^10 atoms (after 5 half-lives)
N = 7.5 x 10^10 atoms / 2 = 3.75 x 10^10 atoms (after 6 half-lives)

We can see that after 6 half-lives (6 x 4.51 billion years = 27.06 billion years), we get to about 3.75 x 10^10 atoms, which is close to the initial number of atoms we calculated earlier. Therefore, the best choice is D, 3.3 x 10^18 atoms.

Question 16

The ¹⁴C activity of some ancient Peruvian corn was found to be 10 disintegrations per minute per gram of carbon. If present-day plant life shows 15 dpm/g, how old is the Peruvian corn? (The half-life of ¹⁴C is 5730 yr.)

Accepted Answer

The age of the Peruvian corn can be calculated using the formula for radioactive decay, which is $N = N_0 \times (1/2)^{t/T}$, where $N$ is the final activity, $N_0$ is the initial activity, $t$ is the time, and $T$ is the half-life. Given that the initial activity (present-day) is 15 dpm/g and the final activity (ancient Peruvian corn) is 10 dpm/g, and knowing the half-life of $^{14}C$ is 5730 years, we can solve for $t$. Rearranging the formula to solve for $t$, we get $t = T \times \log_2(N_0/N)$. Substituting the given values, $t = 5730 \times \log_2(15/10)$, which calculates to approximately 3350 years.

Question 17

Cobalt-60 is a beta emitter with a half-life of 5.3 years. Approximately what fraction of cobalt-60 atoms will remain in a particular sample after 26.5 years?

Accepted Answer

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Question 18

Determine how much energy is released when thorium-230 decays according to the following equation. (1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; C = 2.99792458 × 10⁸ m/s)

Accepted Answer

First, we need to determine the amount of energy released per decay of thorium-230. This can be done by subtracting the mass of the products from the mass of the reactant and converting the mass difference to energy using E=mc^2.

The equation for the decay of thorium-230 is:

^230Th -> ^226Ra + alpha

So,

mass of ^230Th = 230 amu 
mass of ^226Ra = 226 amu 
mass of alpha particle = 4 amu

mass difference = (226 + 4) - 230 = 0 amu 
energy released per decay = (0 amu) x (1 kg / 6.022 × 10^23 amu) x (2.99792458 × 10^8 m/s)^2 = 4.61 × 10^18 J/mol

Finally, we convert this energy to kJ/mol:

4.61 × 10^18 J/mol x (1 kJ / 1000 J) = 4.61 × 10^15 kJ/mol

Rounded to two significant figures, this is 4.61 × 10^18 kJ/mol, which is answer choice B.

Question 19

1 joule equals ________.

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...

Question 20

What is the nuclear binding energy per nucleon of potassium-40? (1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; C = 2.99792458 × 10⁸ m/s)

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Quiz 20: Nuclear Chemistry

Charcoal found under a stone at Stonehenge, England, has a carbon-14 activity that is 0.60 that of new wood. How old is the charcoal? (The half-life of carbon-14 is 5730 years.)

The radiochemist, Will I. Glow, studied thorium-232 and found that 2.82 × 10-7 moles emitted 8.42 × 106 α particles in one year. What is the decay constant for thorium-232?

A mass of 6.02 × 1026 amu is equivalent to

Polonium-208 is an alpha emitter with a half-life of 2.90 years. How many milligrams of polonium from an original sample of 2.00 mg will remain after 8.00 years?

What is the nuclear binding energy per nucleon for ? (1 kg = 6.022 × 1026 amu; C = 2.99792458 × 108 m/s) Particle Mass (amu) 24.985839 1.007276 1.008665

Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation. (1 kg = 6.022 × 1026 amu; NA = 6.022 × 1023 mol-1; C = 2.99792458 × 108 m/s)

Iodine-131, t1/2 = 8.0 days, is used in the diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-131 initially emits 9.95 × 1018 β particles per day, how long will it take for the activity to drop to 6.22 × 1017 β particles per day?

What is the nuclear binding energy per nucleon of uranium-234? [1 kg = 6.022 × 1026 amu; NA = 6.022 × 1023 mol-1; c = 2.99792458 × 108 m/s]

The nuclear binding energy per nucleon is defined by which of the following?

An isotope with a high value of N/Z will tend to decay through

What fraction of radioactive atoms remains in a sample after six half-lives?

Cesium-134 is a β emitter with a half-life of 2.0 years. How much of a 2.50-g sample of cesium-134 will remain after 10 years?

What is the energy equivalent of 1 amu? (c = 2.99792458 × 108 m/s)

What is the name for the difference between the mass of an atom and the sum of the masses of its constituent nucleons?

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half life for 238U → 206Pb is 4.51 × 109 yr.)

The 14C activity of some ancient Peruvian corn was found to be 10 disintegrations per minute per gram of carbon. If present-day plant life shows 15 dpm/g, how old is the Peruvian corn? (The half-life of 14C is 5730 yr.)

Cobalt-60 is a beta emitter with a half-life of 5.3 years. Approximately what fraction of cobalt-60 atoms will remain in a particular sample after 26.5 years?

Determine how much energy is released when thorium-230 decays according to the following equation. (1 kg = 6.022 × 1026 amu; NA = 6.022 × 1023 mol-1; C = 2.99792458 × 108 m/s)

1 joule equals ________.

What is the nuclear binding energy per nucleon of potassium-40? (1 kg = 6.022 × 1026 amu; NA = 6.022 × 1023 mol-1; C = 2.99792458 × 108 m/s)

The radiochemist, Will I. Glow, studied thorium-232 and found that 2.82 × 10^-7 moles emitted 8.42 × 10⁶ α particles in one year. What is the decay constant for thorium-232?

A mass of 6.02 × 10²⁶ amu is equivalent to

What is the nuclear binding energy per nucleon for ? (1 kg = 6.022 × 10²⁶ amu; C = 2.99792458 × 10⁸ m/s) Particle Mass (amu) 24.985839 1.007276 1.008665

Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation. (1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; C = 2.99792458 × 10⁸ m/s)

Iodine-131, t_1/2= 8.0 days, is used in the diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-131 initially emits 9.95 × 10¹⁸ β particles per day, how long will it take for the activity to drop to 6.22 × 10¹⁷ β particles per day?

What is the nuclear binding energy per nucleon of uranium-234? [1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; c = 2.99792458 × 10⁸ m/s]

What is the energy equivalent of 1 amu? (c = 2.99792458 × 10⁸ m/s)

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half life for ²³⁸U → ²⁰⁶Pb is 4.51 × 10⁹ yr.)

The ¹⁴C activity of some ancient Peruvian corn was found to be 10 disintegrations per minute per gram of carbon. If present-day plant life shows 15 dpm/g, how old is the Peruvian corn? (The half-life of ¹⁴C is 5730 yr.)

Determine how much energy is released when thorium-230 decays according to the following equation. (1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; C = 2.99792458 × 10⁸ m/s)

What is the nuclear binding energy per nucleon of potassium-40? (1 kg = 6.022 × 10²⁶ amu; N_A = 6.022 × 10²³ mol^-1; C = 2.99792458 × 10⁸ m/s)