Question 1

A sample of parchment produces 10.3 disintegrations of ¹⁴C per minute per gram of carbon. Estimate the age of the parchment, assuming the original activity was 15.3 disintegrations per minute per gram of carbon. (t_1/2 = 5730 years)

Accepted Answer

The age can be estimated using the formula for radioactive decay: N(t) = N0 * (1/2)^(t/t1/2). Solving for t gives approximately 3,300 years.

Question 2

The half-life of Nitrogen - 16 is 7 seconds. How long does it take for 100 mg of Nitrogen - 16 to be reduced tO₆.25 mg?

Accepted Answer

The half-life of Nitrogen-16 is 7 seconds, meaning it takes 7 seconds for the amount to reduce to half. Starting with 100 mg, it reduces to 50 mg in 7 seconds, 25 mg in 14 seconds, 12.5 mg in 21 seconds, and 6.25 mg in 28 seconds.

Question 3

Iodine - 131 is a radionuclide that is frequently used in nuclear medicine. Among other things, it is used to detect fluid buildup in the brain. The half-life of iodine - 131 is 8.0 days. After introducing this radionuclide to a patient, if only 1/16^th of the original sample remains, how many days have expired?

Accepted Answer

The answer of Iodine - 131 is a radionuclide that...

Question 4

Phosphorous - 32 has a half-life of 14.3 days. How many days are required for 4.00 grams of phosphorous - 32 to decay to 1.00 gram?

Accepted Answer

The decay of a substance to a quarter of its original amount requires two half-lives. Since the half-life of phosphorous-32 is 14.3 days, two half-lives would be 28.6 days.

Question 5

Technetium - 99 has a half-life equal tO₆.0 hours. What fraction of Technetiun - 99 remains undecayed after 18 hours?

Accepted Answer

We can use the formula for radioactive decay:

N = N0 * (1/2)^(t/t1/2)

where N is the remaining amount, N0 is the initial amount, t is the time elapsed, and t1/2 is the half-life.

Plugging in the given values, we get:

N = N0 * (1/2)^(18/6.02) = N0 * (1/2)^3 = N0 * 1/8

This means that only 1/8 of the Technetium-99 remains undecayed after 18 hours, so the answer is C.

Question 6

What type of nuclear reaction is observed in the following process? ²⁴⁸₉₆Cm + ²⁸₁₀Ne→¹¹⁶₅₁Sb + ¹⁶⁰₅₅Cs

Accepted Answer

The presence of Cm and Ne indicates that this is a nuclear fission reaction.

Question 7

What type of nuclear reaction is illustrated below? ²³⁵₉₂U + ¹₀n→¹³⁹₅₄Xe + ⁹⁵₃₈Sr + 2 ¹₀n

Accepted Answer

The presence of a neutron in the reactants, and the fact that it splits into two different products (Xe and Sr) suggest that this is a nuclear fission reaction, in which a heavy nucleus (in this case, likely Uranium-235) is split into two lighter nuclei. The emission of neutrons during this process is also characteristic of nuclear fission.

Question 8

The half-life of ¹⁴C is 5730 years. Carbon - 14 dating would not be useful in the dating of which of the following samples?

Accepted Answer

Carbon-14 dating can only be used to determine the age of organic materials up to approximately 50,000 years old, as the half-life of carbon-14 is only 5730 years. Paper from the 1970s would not be old enough to use carbon-14 dating, as its age is easily determinable through other means. Therefore, choice C would not be useful for carbon-14 dating.

Question 9

The reaction below can be best described as what type of nuclear reaction? ²³⁵₉₂U + ¹₀n→¹³⁹₅₆Ba + ⁹⁴₃₆Kr + 3 ¹₀n

Accepted Answer

The reaction involves the splitting of a nucleus into two or more smaller nuclei, which is characteristic of nuclear fission.

Question 10

Which of the following types of nuclear reactions is(are) considered bombardment reactions ?
I. Radioactive decay
II. Nuclear fission
III. Nuclear fusion

Accepted Answer

Bombardment reactions involve bombarding a nucleus with a particle to induce a nuclear reaction. Nuclear fission and fusion both involve bombarding a nucleus with a particle (usually a neutron) to induce a nuclear reaction. Radioactive decay is not a bombardment reaction as it does not involve bombarding a nucleus with a particle.

Question 11

Barium - 122 has a half-life of 2 minutes. You just obtained a sample weighing 10.0 grams. It takes 10 minutes to set up the experiment in which barium - 122 will be used. How many grams of barium - 122 will be left when you begin the experiment?

Accepted Answer

Using the formula for radioactive decay, we can calculate the amount of barium - 122 remaining after 10 minutes:

N = N0 * (1/2)^(t/h)

where:
N0 = initial amount of barium - 122 = 10.0 g
N = final amount of barium - 122 (what we're solving for)
t = time elapsed = 10 min
h = half-life = 2 min

Plugging in the values:

N = 10.0 g * (1/2)^(10/2) = 10.0 g * (1/2)^5 = 10.0 g * 0.03125 = 0.313 g

Therefore, the answer is B, 0.313 g of barium - 122 will be left when the experiment begins.

Question 12

In a nuclear reaction, the nuclide ⁷⁵As absorbs a neutron and emits a gamma ray. The product nucleus is:

Accepted Answer

When a nucleus absorbs a neutron and emits a gamma ray, its mass number increases by one (due to the addition of a neutron) but its atomic number remains unchanged (since a gamma ray does not alter the number of protons). Therefore, the product nucleus will have the same atomic number but a mass number increased by one, making option D ("⁷⁶As") the correct answer.

Question 13

A Uranium ore contains 0.250 g of ²⁰⁶Pb per gram of Uranium. Assuming all of the lead was produced by decay of ²³⁸U (t_1/2 = 4.5×10⁹ years), what is the age of the rock?

Accepted Answer

The answer of A Uranium ore contains 0.250 g of...

Question 14

A 4.00 gram sample of a radionuclide at room temperature has a half-life of 91.3 days. How much of the sample remains after 365 days?

Accepted Answer

The amount of radionuclide remaining after a given time can be calculated using the formula for exponential decay, which is $A = A_0 \times (1/2)^{t/T}$, where $A$ is the amount remaining, $A_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life. Plugging in the given values: $4.00 \times (1/2)^{365/91.3} \approx 0.250$ grams.

Question 15

The half-life of strontium - 90 is 28 years. How many grams of this nuclide in a 4.00 gram sample will remain after 112 years?

Accepted Answer

The half-life of strontium-90 is 28 years, so after 112 years (which is 4 half-lives: 112 years / 28 years/half-life = 4 half-lives), the amount of strontium-90 will have halved 4 times. Starting with 4.00 grams, after one half-life, it would be 2.00 grams, then 1.00 gram, then 0.50 gram, and finally 0.25 gram after the fourth half-life.

Question 16

Assuming that the original activity of carbon - 14 was 15.3 disintegrations/min/g, what is the present disintegration rate (in disintegrations/min/g) in a sample of parchment that is 2.10×10³ years old? (t_1/2 = 5730 years)

Accepted Answer

We use the formula for radioactive decay:

N(t) = N0 * (1/2)^(t/t1/2)

Where:
N(t) = the number of nuclei at time t
N0 = the initial number of nuclei
t1/2 = the half-life
t = elapsed time

We can rewrite this as:

N(t) = N0 * e^(-λt)

Where λ is the decay constant, given by:

λ = ln(2)/t1/2

Let's plug in the values we have:

t = 2.10×10^13 years
t1/2 = 5730 years
N0/N(t) = 1/2

We want to solve for the present disintegration rate, which is given by λN(t).

First, let's solve for N/N0:

1/2 = (1/2)^(t/t1/2)
1/2 = e^(-λt)
ln(1/2) = -λt
λ = ln(2)/t1/2 = ln(2)/5730 years
t = 2.10×10^13 years = 2.10×10^13 years / (365.25 days/year) / (24 hours/day) / (60 minutes/hour) = 3.98×10^20 minutes

Now we can solve for N/N0:

N(t)/N0 = e^(-λt) = e^(-(ln2/5730 years)*(3.98×10^20 minutes))
N(t)/N0 = 0.0123

So the present disintegration rate is:

λN(t) = (ln(2)/5730 years) * 15.3 disintegrations/min/g * 0.0123
λN(t) = 0.000124 disintegrations/min/g

Rounding to one significant figure, we get:

λN(t) = 0.0001 disintegrations/min/g ≈ 0.0001 ≈ 1.2×10^-4 ≈ 0.00012 ≈ 0.00013

The closest answer is B) 11.9.

Question 17

What type of nuclear reaction is illustrated in the following transformation? ²³⁵₉₂U + ¹₀n→¹⁴⁴₅₅Cs + ⁹⁰₃₇Rb + 2 ¹₀n

Accepted Answer

This transformation shows a heavy nucleus (uranium) splitting into smaller nuclei (cesium and rubidium) along with the release of additional neutrons, which is characteristic of nuclear fission.

Question 18

What type of nuclear reaction is illustrated below? ³₁H + ²₁H→⁴₂He + ¹₀n

Accepted Answer

This reaction involves the fusion of two hydrogen nuclei (protons) to form helium and a neutron, which is a type of nuclear fusion reaction.

Question 19

Cobalt - 60 has a half-life equal tO₅ .3 years. What fraction of cobalt - 60 remains undecayed after 21.2 years?

Accepted Answer

To find the fraction of Cobalt-60 that remains after 21.2 years, we use the formula for radioactive decay, which involves the half-life of the substance. The half-life of Cobalt-60 is 5.3 years. To find out how many half-lives have passed in 21.2 years, we divide 21.2 by 5.3, which gives us exactly 4 half-lives. After each half-life, half of the substance remains, so after 4 half-lives, we calculate the remaining fraction as (1/2)^4 = 1/16 = 0.0625.

Question 20

Polonium - 218, a decay product of radon - 222, has a half-life of 3 minutes. What percentage of the polonium - 218 formed will remain in the lung 9 minutes after inhalation?

Accepted Answer

After 9 minutes, which is three half-lives of Polonium-218 (3 minutes per half-life), the amount remaining would be (1/2)^3 = 1/8, or 12.5% of the original amount.

Quiz 21: Nuclear Chemistry

A sample of parchment produces 10.3 disintegrations of ¹⁴C per minute per gram of carbon. Estimate the age of the parchment, assuming the original activity was 15.3 disintegrations per minute per gram of carbon. (t_1/2 = 5730 years)

The half-life of Nitrogen - 16 is 7 seconds. How long does it take for 100 mg of Nitrogen - 16 to be reduced tO₆.25 mg?

Phosphorous - 32 has a half-life of 14.3 days. How many days are required for 4.00 grams of phosphorous - 32 to decay to 1.00 gram?

Technetium - 99 has a half-life equal tO₆.0 hours. What fraction of Technetiun - 99 remains undecayed after 18 hours?

What type of nuclear reaction is observed in the following process? ²⁴⁸₉₆Cm + ²⁸₁₀Ne→¹¹⁶₅₁Sb + ¹⁶⁰₅₅Cs

What type of nuclear reaction is illustrated below? ²³⁵₉₂U + ¹₀n→¹³⁹₅₄Xe + ⁹⁵₃₈Sr + 2 ¹₀n

The half-life of ¹⁴C is 5730 years. Carbon - 14 dating would not be useful in the dating of which of the following samples?

The reaction below can be best described as what type of nuclear reaction? ²³⁵₉₂U + ¹₀n→¹³⁹₅₆Ba + ⁹⁴₃₆Kr + 3 ¹₀n

Which of the following types of nuclear reactions is(are) considered bombardment reactions ? I. Radioactive decay II. Nuclear fission III. Nuclear fusion

Barium - 122 has a half-life of 2 minutes. You just obtained a sample weighing 10.0 grams. It takes 10 minutes to set up the experiment in which barium - 122 will be used. How many grams of barium - 122 will be left when you begin the experiment?

In a nuclear reaction, the nuclide ⁷⁵As absorbs a neutron and emits a gamma ray. The product nucleus is:

A Uranium ore contains 0.250 g of ²⁰⁶Pb per gram of Uranium. Assuming all of the lead was produced by decay of ²³⁸U (t_1/2 = 4.5×10⁹ years) , what is the age of the rock?

A 4.00 gram sample of a radionuclide at room temperature has a half-life of 91.3 days. How much of the sample remains after 365 days?

The half-life of strontium - 90 is 28 years. How many grams of this nuclide in a 4.00 gram sample will remain after 112 years?

Assuming that the original activity of carbon - 14 was 15.3 disintegrations/min/g, what is the present disintegration rate (in disintegrations/min/g) in a sample of parchment that is 2.10×10³ years old? (t_1/2 = 5730 years)

What type of nuclear reaction is illustrated in the following transformation? ²³⁵₉₂U + ¹₀n→¹⁴⁴₅₅Cs + ⁹⁰₃₇Rb + 2 ¹₀n

What type of nuclear reaction is illustrated below? ³₁H + ²₁H→⁴₂He + ¹₀n

Cobalt - 60 has a half-life equal tO₅ .3 years. What fraction of cobalt - 60 remains undecayed after 21.2 years?

Polonium - 218, a decay product of radon - 222, has a half-life of 3 minutes. What percentage of the polonium - 218 formed will remain in the lung 9 minutes after inhalation?

Quiz 21: Nuclear Chemistry

A sample of parchment produces 10.3 disintegrations of 14C per minute per gram of carbon. Estimate the age of the parchment, assuming the original activity was 15.3 disintegrations per minute per gram of carbon. (t1/2 = 5730 years)

The half-life of Nitrogen - 16 is 7 seconds. How long does it take for 100 mg of Nitrogen - 16 to be reduced tO6.25 mg?

Phosphorous - 32 has a half-life of 14.3 days. How many days are required for 4.00 grams of phosphorous - 32 to decay to 1.00 gram?

Technetium - 99 has a half-life equal tO6.0 hours. What fraction of Technetiun - 99 remains undecayed after 18 hours?

What type of nuclear reaction is observed in the following process? 24896Cm + 2810Ne→11651Sb + 16055Cs

What type of nuclear reaction is illustrated below? 23592U + 10n→13954Xe + 9538Sr + 2 10n

The half-life of 14C is 5730 years. Carbon - 14 dating would not be useful in the dating of which of the following samples?

The reaction below can be best described as what type of nuclear reaction? 23592U + 10n→13956Ba + 9436Kr + 3 10n

Which of the following types of nuclear reactions is(are) considered bombardment reactions ? I. Radioactive decay II. Nuclear fission III. Nuclear fusion

Barium - 122 has a half-life of 2 minutes. You just obtained a sample weighing 10.0 grams. It takes 10 minutes to set up the experiment in which barium - 122 will be used. How many grams of barium - 122 will be left when you begin the experiment?

In a nuclear reaction, the nuclide 75As absorbs a neutron and emits a gamma ray. The product nucleus is:

A Uranium ore contains 0.250 g of 206Pb per gram of Uranium. Assuming all of the lead was produced by decay of 238U (t1/2 = 4.5×109 years) , what is the age of the rock?

A 4.00 gram sample of a radionuclide at room temperature has a half-life of 91.3 days. How much of the sample remains after 365 days?

The half-life of strontium - 90 is 28 years. How many grams of this nuclide in a 4.00 gram sample will remain after 112 years?

Assuming that the original activity of carbon - 14 was 15.3 disintegrations/min/g, what is the present disintegration rate (in disintegrations/min/g) in a sample of parchment that is 2.10×103 years old? (t1/2 = 5730 years)

What type of nuclear reaction is illustrated in the following transformation? 23592U + 10n→14455Cs + 9037Rb + 2 10n

What type of nuclear reaction is illustrated below? 31H + 21H→42He + 10n

Cobalt - 60 has a half-life equal tO5 .3 years. What fraction of cobalt - 60 remains undecayed after 21.2 years?

Polonium - 218, a decay product of radon - 222, has a half-life of 3 minutes. What percentage of the polonium - 218 formed will remain in the lung 9 minutes after inhalation?

A sample of parchment produces 10.3 disintegrations of ¹⁴C per minute per gram of carbon. Estimate the age of the parchment, assuming the original activity was 15.3 disintegrations per minute per gram of carbon. (t_1/2 = 5730 years)

The half-life of Nitrogen - 16 is 7 seconds. How long does it take for 100 mg of Nitrogen - 16 to be reduced tO₆.25 mg?

Technetium - 99 has a half-life equal tO₆.0 hours. What fraction of Technetiun - 99 remains undecayed after 18 hours?

What type of nuclear reaction is observed in the following process? ²⁴⁸₉₆Cm + ²⁸₁₀Ne→¹¹⁶₅₁Sb + ¹⁶⁰₅₅Cs

What type of nuclear reaction is illustrated below? ²³⁵₉₂U + ¹₀n→¹³⁹₅₄Xe + ⁹⁵₃₈Sr + 2 ¹₀n

The half-life of ¹⁴C is 5730 years. Carbon - 14 dating would not be useful in the dating of which of the following samples?

The reaction below can be best described as what type of nuclear reaction? ²³⁵₉₂U + ¹₀n→¹³⁹₅₆Ba + ⁹⁴₃₆Kr + 3 ¹₀n

In a nuclear reaction, the nuclide ⁷⁵As absorbs a neutron and emits a gamma ray. The product nucleus is:

A Uranium ore contains 0.250 g of ²⁰⁶Pb per gram of Uranium. Assuming all of the lead was produced by decay of ²³⁸U (t_1/2 = 4.5×10⁹ years) , what is the age of the rock?

Assuming that the original activity of carbon - 14 was 15.3 disintegrations/min/g, what is the present disintegration rate (in disintegrations/min/g) in a sample of parchment that is 2.10×10³ years old? (t_1/2 = 5730 years)

What type of nuclear reaction is illustrated in the following transformation? ²³⁵₉₂U + ¹₀n→¹⁴⁴₅₅Cs + ⁹⁰₃₇Rb + 2 ¹₀n

What type of nuclear reaction is illustrated below? ³₁H + ²₁H→⁴₂He + ¹₀n

Cobalt - 60 has a half-life equal tO₅ .3 years. What fraction of cobalt - 60 remains undecayed after 21.2 years?