Question 1

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 activity decreases from 1000 to 125 decays/sec, which is a reduction by a factor of 8 (1000 to 500 to 250 to 125). This corresponds to three half-lives (since 1000/2 = 500, 500/2 = 250, 250/2 = 125). Since this occurs over 3 hours, the half-life is 1 hour.

Question 2

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

Accepted Answer

The number of atoms can be calculated using the formula N = (Activity x Half-life) / (0.693 x Decay constant). Converting the activity to decays per second and using the half-life, the calculation yields approximately 8 x 10^12 atoms.

Question 3

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

Accepted Answer

The mass of an atom is not simply the sum of the masses of its protons, neutrons, and electrons. Rather, the mass is slightly less than the sum due to the binding energy that holds the nucleus together. Uranium, in particular, has a large binding energy due to its high atomic number, which causes the mass of an individual uranium atom to be slightly less than an integer multiple of the mass of a hydrogen atom.

Question 4

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 x 10¹⁸ tritium atoms,what is its activity in decays/second? (1 year = 3.15 x 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 5

A pure sample of ²²⁶Ra contains 2.0 x 10¹⁴atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 x 10³years,what is the decay rate of this sample? (1 Ci = 3.7 x 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 6

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

Accepted Answer

The decay constant (lambda) is calculated using the equation: lambda = ln(2)/t_half, where ln represents natural logarithm. Substituting the given value of half-life (t_half = 0.62 s) in this equation, we get: lambda = ln(2)/0.62 s ≈ 1.1 s^-1. Therefore, the correct choice is B. Option A and C can be eliminated because they provide different values for lambda. Option D is incorrect as the decay constant can still be calculated for a given half-life, regardless of its magnitude.

Question 7

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 x 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 8

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

Accepted Answer

If 15% of the radioactive material decays in 3 hours, then we can assume that 30% will decay in 6 hours (since half of the remaining material will decay in the next 3 hours). Similarly, 60% will decay in 9 hours, and 90% will decay in 12 hours.

Therefore, the material's half-life can be calculated by solving the following equation:

0.5 = (1 - 0.15)^n

Where n is the number of half-lives.

Solving for n, we get:

n = log(0.5) / log(0.85)

n ≈ 4.089

Multiplying by 3 hours (the length of one half-life), we get:

t1/2 ≈ 12.27 hours

So the closest answer choice is B) 12.8 hrs.

Question 9

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 10

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 particles are the least penetrating, requiring minimal shielding, typically a sheet of paper or the outer layers of human skin can stop them. Beta particles are more penetrating than alpha particles but can be stopped by a few millimeters of aluminum. Gamma rays are the most penetrating and require dense materials like lead or several centimeters of concrete for effective shielding.

Question 11

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 fossil contains 1/8th of the original Carbon-14, which means it has gone through 3 half-lives (since 1/2^3 = 1/8). Each half-life is 5,730 years, so 3 half-lives is 3 * 5,730 = 17,190 years, approximately 17,200 years.

Question 12

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 13

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 directly proportional to its half-life. So, if the half-life of X is twice that of Y, then it will decay at a slower rate than Y. Therefore, X has a smaller decay rate than Y.

Question 14

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

The activity of a radioactive isotope sample reduces by a factor of two in each half-life period. Therefore, to reduce the activity to 0.004 of the original value, we need to divide the original activity by 2 six times:

1/2^6 = 1/64 ≈ 0.016

So, approximately 6 half-life periods must elapse for the activity of the sample to be reduced to 0.004 of the original value. Therefore, the correct answer is choice C.

Question 15

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 16

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

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

Question 17

1 Bq = ________ Ci?

Accepted Answer

1 Bq equals 2.7 x 10^-11 Ci. Option A is incorrect because 1 Bq is not equal to 1 Ci. Option B is a combination of characters and does not make any sense. Option D is incorrect because the value given is too large.

Question 18

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

Accepted Answer

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

Question 19

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

Accepted Answer

The binding energy per nucleon being relatively constant despite the volume of the nucleus increasing indicates that the strong nuclear force saturates, meaning that it has a maximum strength that it can exert on nucleons within the nucleus. As more and more nucleons are added to the nucleus, the force between them becomes increasingly crowded and the total energy of the nucleus becomes less favorable. This saturation of the strong force leads to the plateau in the binding energy per nucleon observed in stable nuclei.  Therefore, the best choice is (A) the strong nuclear force saturates.

Question 20

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.

Quiz 29: Nuclear Physics

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?

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

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

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 x 1018 tritium atoms,what is its activity in decays/second? (1 year = 3.15 x 107 s)

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

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

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 x 1010 decays/second)