Question 1

Tritium is radioactive with a half-life of 12.33 years decaying into 3He with low-energy electron emission.If we have a sample of 3.00 * 10¹⁸ tritium atoms, what is its activity in decays/second? (1 year = 3.15 * 10⁷ s)

Accepted Answer

First, we need to calculate the decay constant, λ, using the half-life given:

t1/2 = 12.33 years = 12.33 x 3.15 x 10^7 s/year ≈ 3.89 x 10^8 s

λ = ln2 / t1/2 ≈ 1.78 x 10^-9 s^-1

The activity, A, is defined as the number of decays per second, so we can use the following equation:

A = λN

where N is the number of radioactive nuclei. Substituting the given values, we get:

A = (1.78 x 10^-9 s^-1) x (3.00 x 10^21 nuclei) ≈ 5.35 x 10^9 decays/second

Therefore, the best choice is B, 5.35 x 10^19/second.

Question 2

There are samples of two different isotopes, X and Y.Both contain the same number of radioactive atoms.Sample X has a half-life twice that of Y.How do their decay rates compare?

Accepted Answer

The decay rate of a radioactive substance is inversely proportional to its half-life. This means that the shorter the half-life, the faster the decay rate. Therefore, if the half-life of sample X is twice that of Y, then sample Y has a shorter half-life, which means it has a faster decay rate than sample X. So, sample X has a smaller decay rate than Y. Therefore, the correct choice is B.

Question 3

The half-life of ¹⁸N is 0.62 s. What is the decay constant for this isotope?

Accepted Answer

The decay constant is calculated using the formula lambda = ln(2)/t_half. Plugging in the given half-life, we get lambda = ln(2)/0.62 s = 1.1 s^-1. The correct answer is therefore B.

Question 4

A pure sample of ²²⁶Ra contains 2.0 * 10¹⁴atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 * 10³years, what is the decay rate of this sample? (1 Ci = 3.7 * 10¹⁰ decays/s)

Accepted Answer

First, we need to find the decay constant (λ) using the formula: 
λ = ln(2)/t1/2 = ln(2)/(1.6 x 10^13 s) = 4.34 x 10^-14 s^-1

Then, we can find the activity (A) using the formula: 
A = λN = (4.34 x 10^-14 s^-1)(2.0 x 10^14 atoms) = 8.68 x 10^6 decays/s

Finally, we can convert the activity to curies using the conversion factor: 
1 Ci = 3.7 x 10^10 decays/s
So, 
A = 8.68 x 10^6 decays/s = 8.68 x 10^6/3.7 x 10^10 Ci = 2.35 x 10^-4 Ci

Therefore, the decay rate of the sample is 2.35 x 10^-4 Ci, which is closest to choice C (7.4 x 10^-8 Ci).

Question 5

A radioactive material initially is observed to have an activity of 1 000 decays/sec.If three hours later it is observed to have an activity of 125 decays/sec, what is its half-life?

Accepted Answer

The answer of A radioactive material initially is observed to...

Question 6

Approximately how many half-life periods must elapse if the activity of a radioactive isotope sample is to be reduced to 0.004 of the original value?

Accepted Answer

If the activity of a radioactive isotope is reduced to 0.004 of the original value, then the fraction remaining is 1/250 of its original value. This means that it has undergone 8 half-life periods, because (1/2)^8 = 1/256, which is just slightly greater than 1/250. Therefore, the correct choice is C, 8.

Question 7

If the stable nuclei are plotted with neutron number vs.proton number, the curve formed by the stable nuclei does not follow the line N = Z.This is predicted by examining how the binding energy is influenced by:

Accepted Answer

The Coulomb repulsion between positively charged protons is a significant factor that affects the stability of the nucleus. As the number of protons increases, the electrostatic repulsion also increases, causing the nucleus to become less stable. This effect is not accounted for by the N = Z line and explains why the stable nuclei curve deviates from it. The other factors (A, B, D) also have an impact on the binding energy of the nucleus but are not the primary reason for the deviation from the N = Z line.

Question 8

The half-life of radioactive Technetium-99 is 6.0 hours.Find the number of ⁹⁹Tc nuclei necessary to produce a sample of activity 1.0 µCi.(1 Ci = 3.7 * 10¹⁰ decays/second)

Accepted Answer

The activity of a radioactive sample is given by:

Activity = Decay constant x Number of radioactive nuclei

Since the activity is given as 1.0 µCi = 3.7 x 10^7 decays/second, and the decay constant for Technetium-99 is given as λ = ln(2) / half-life = ln(2) / 6.0 hours = 0.1155 hours^-1, we can solve for the number of radioactive nuclei:

1.0 µCi = (3.7 x 10^7 decays/second) = (0.1155 hours^-1) x N
N = (1.0 µCi) / (3.7 x 10^7 decays/second) x (1/0.1155 hours^-1)
N = 1.2 x 10^19 nuclei

Therefore, the answer is choice B: 1.2 x 10^19.

Question 9

The fact that the binding energy per nucleon does not depend very strongly on the volume of the nucleus indicates that:

Accepted Answer

The fact that the binding energy per nucleon does not depend very strongly on the volume of the nucleus is an indication that the strong nuclear force saturates. This means that the force becomes constant beyond a certain range of distances, which limits the total binding energy per nucleon. Therefore, the binding energy per nucleon reaches a maximum value and then levels off as the number of nucleons in the nucleus increases.

Question 10

A pure sample of ²²⁶Ra contains 2.0 * 10¹⁴atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 * 10³ years, what is the activity of this sample?

Accepted Answer

The answer of A pure sample of ²²⁶Ra contains 2.0...

Question 11

Tritium has a half-life of 12.3 years.How many years will elapse when the radioactivity of a tritium sample diminishes to 20% of its original value?

Accepted Answer

To solve this problem, we can use the half-life formula:
N(t) = N₀(1/2)^(t/T)
where:
N(t) = the amount at time t
N₀ = the initial amount
t = time elapsed
T = the half-life

We are given that T = 12.3 years and we want to find the time elapsed when N(t) = 0.2*N₀. So we can set up the equation:
0.2*N₀ = N₀(1/2)^(t/12.3)
Dividing both sides by N₀ gives:
0.2 = (1/2)^(t/12.3)
Taking the logarithm of both sides:
log(0.2) = (t/12.3)*log(1/2)
Simplifying:
t/12.3 = log(0.2)/log(1/2)
t = 12.3*(log(0.2)/log(1/2))
t ≈ 29 years

Therefore, the best choice is B) 29 years.

Question 12

An element is emitting alpha, beta, and gamma radiation.Rank order them in terms of the thickness of protective shielding they will need for safety, from least to most.

Accepted Answer

Alpha radiation can be easily stopped by a sheet of paper or human skin, so it requires the least amount of protective shielding. Beta radiation is a bit more penetrating and can be stopped by a layer of clothing or a few centimeters of a non-metallic material, so it requires slightly more protection than alpha radiation. Gamma radiation, on the other hand, is highly penetrating and requires several centimeters of dense material (like lead or concrete) to stop it. Therefore, alpha requires the least protective shielding and gamma requires the most, and the correct order is alpha, beta, and gamma.

Question 13

Calculate the binding energy per nucleon of the tritium nucleus, , given that the mass of the tritium nucleus is 3.016 05 u.(m_p = 1.007 276 u, mn = 1.008 665, and 1 u = 931.5 MeV/c²)

Accepted Answer

The answer of Calculate the binding energy per nucleon of...

Question 14

1 Bq = ________ Ci?

Accepted Answer

1 Becquerel (Bq) is equal to 2.7 * 10^-11 Curie (Ci).

Question 15

The mass of ²³⁸U is not quite an integer multiple of ¹H mass.This is primarily because:

Accepted Answer

The mass of an atomic nucleus is slightly less than the sum of the masses of its individual protons and neutrons. This is due to the binding energy of the nucleus, which is the energy required to split the nucleus into its constituent parts. Uranium has a particularly high binding energy due to its large number of protons and neutrons, which means that its actual mass is less than what would be expected based on the masses of its constituent particles. This is why the mass of ²³⁸U is not quite an integer multiple of ¹H mass.

Question 16

An ancient building was known to have been built 3 000 years ago.Approximately what proportion of Carbon-14 atoms are yet in the building's wooden framing compared to the number which were present at the time of its construction? (half life of ¹⁴C = 5 730 years)

Accepted Answer

The answer of An ancient building was known to have...

Question 17

If a fossil bone is found to contain 1/8th as much Carbon-14 as the bone of a living animal, what is the approximate age of the fossil? (half-life of ¹⁴C = 5 730 years)

Accepted Answer

The answer of If a fossil bone is found to...

Question 18

Tritium has a half-life of 12.3 years.What proportion of its original radioactivity will a sample have after 9 years?

Accepted Answer

The answer of Tritium has a half-life of 12.3 years.What...

Question 19

Approximately how many radioactive atoms are present in a tritium sample with an activity of 0.4 * 10^-6 Ci and a half-life of 12.3 years? (1 Ci = 3.7 * 10¹⁰ decays/s)

Accepted Answer

The answer of Approximately how many radioactive atoms are present...

Question 20

Over the course of 3 hours, 15% of a radioactive material decays.What is its half-life?

Accepted Answer

If 15% of material decays in 3 hours, then we can calculate the initial amount of material as follows:
Let N be the initial amount of material.
After 1 half-life, the amount of material remaining will be N/2
After 2 half-lives, the amount of material remaining will be (N/2)/2 = N/4
After 3 half-lives, the amount of material remaining will be (N/2)/2)/2 = N/8
We know that 15% of material has decayed after 3 hours, so the amount of material remaining is 85% of the initial amount.
Thus, we have:
85%N = N/8
Solving for N, we get:
N = (8/0.85)N = 9.4118N
So, the time for one half-life can be calculated as follows:
N/2 = 4.7059N
Since 4.7059N is the amount of material remaining after one half-life, and we know that 15% of material has decayed after 3 hours, we can set up the following equation:
4.7059N = 0.85N
Solving for the time, we get:
t = 3 hours / (ln 2) = 4.1 hours
Therefore, the half-life is approximately 12.8 hours.

Quiz 29: Nuclear Physics

Tritium is radioactive with a half-life of 12.33 years decaying into 3He with low-energy electron emission.If we have a sample of 3.00 * 1018 tritium atoms, what is its activity in decays/second? (1 year = 3.15 * 107 s)

There are samples of two different isotopes, X and Y.Both contain the same number of radioactive atoms.Sample X has a half-life twice that of Y.How do their decay rates compare?

The half-life of 18N is 0.62 s. What is the decay constant for this isotope?

A pure sample of 226Ra contains 2.0 * 1014 atoms of the isotope.If the half-life of 226Ra = 1.6 * 103 years, what is the decay rate of this sample? (1 Ci = 3.7 * 1010 decays/s)