Question 1

Tritium has a half-life of 12.3 years.How many years will elapse when the radioactivity of a tritium sample diminishes to 35% of its original value?
A)1.2E-1 years
B)1.9E+1 years
C)4.3E+0 years
D)6.3E+1 years
E)​8.1E+0 years

Accepted Answer

The half-life formula is used to calculate the time it takes for a quantity to reduce to half its initial value. The formula is $ t = \frac{\ln(\frac{final\ amount}{initial\ amount})}{\ln(0.5)} \times half\ life $. Substituting the given values, $ t = \frac{\ln(0.35)}{\ln(0.5)} \times 12.3 $ years, which calculates to approximately 19 years.

Question 2

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
A)the volume of the nucleus.
B)the size of the nuclear surface.
C)the Coulomb repulsion.
D)the proton-neutron mass difference.

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 3

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) A)0.425 B)0.500 C)0.517 D)0.696

Accepted Answer

The proportion of Carbon-14 remaining can be calculated using the formula (1/2)^(t/half-life). For 3,000 years, it is approximately (1/2)^(3000/5730) ≈ 0.696.

Question 4

The half-life of radioactive Technetium-99 is 6.0 hours.Find the number of ⁹⁹Tc nuclei necessary to produce a sample of activity 1.9 µCi.(1 Ci = 3.7 × 10¹⁰ decays/second) A)5.9E-2 B)2.2E+9 C)1.5E+9 D)6.1E+5 E)4.2E+5

Accepted Answer

The activity (A) of a radioactive sample is given by the formula $A = \lambda N$, where $A$ is the activity in decays per second, $\lambda$ is the decay constant, and $N$ is the number of radioactive nuclei. The decay constant $\lambda$ can be calculated from the half-life ($T_{1/2}$) using the formula $\lambda = \frac{\ln(2)}{T_{1/2}}$. Given that the half-life of Technetium-99 is 6.0 hours, we first convert this to seconds: $6.0 \, \text{hours} \times 3600 \, \text{seconds/hour} = 21600 \, \text{seconds}$. Then, we calculate the decay constant: $\lambda = \frac{\ln(2)}{21600} \, \text{seconds}^{-1}$.Given an activity of 1.9 µCi, we convert this to decays per second: $1.9 \, \mu\text{Ci} \times 3.7 \times 10^{10} \, \text{decays/second/Ci} = 7.03 \times 10^4 \, \text{decays/second}$.Now, we can solve for $N$: $N = \frac{A}{\lambda} = \frac{7.03 \times 10^4}{\ln(2)/21600}$.Plugging in the values, we find $N \approx 2.2 \times 10^9$, which corresponds to choice B.

Question 5

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?
A)X has a greater rate than Y.
B)X has a smaller rate than Y.
C)The rates of X and Y are equal.
D)The rate depends on atomic number,not half-life.

Accepted Answer

The decay rate of a radioactive isotope is related to its half-life, where a shorter half-life implies a faster decay rate. Since sample X has a half-life twice that of Y, it will decay slower than Y. Therefore, option B is correct, and X has a smaller decay rate compared to Y.

Question 6

The mass of ²³⁸U is not quite an integer multiple of ¹H mass.This is primarily because A)the proton and neutron have different masses. B)there are several isotopes of uranium. C)of the binding energy of uranium. D)uranium is radioactive.

Accepted Answer

The mass of an atomic nucleus is less than the combined mass of its individual protons and neutrons because of the binding energy that holds the nucleus together. The binding energy per nucleon is different for different elements, and uranium has a particularly high binding energy per nucleon. This means that the mass of a uranium nucleus is significantly less than the sum of the masses of its individual protons and neutrons. Therefore, the mass of ²³⁸U is not quite an integer multiple of ¹H mass.

Question 7

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 1.91 × 10¹⁸ tritium atoms,what is its activity in decays/second? (1 year = 3.15 × 10⁷ s) A)4.92E+9 decays/second B)3.41E+9 decays/second C)1.07E+17 decays/second D)6.69E+16 decays/second E)3.41E+10 decays/second

Accepted Answer

First, we need to calculate the lambda (decay constant) for tritium:

lambda = ln(2)/half-life = ln(2)/12.33 years = 0.0563/year

Next, we need to convert the sample size from atoms to moles:

n = 1.91 x 10^18 atoms / 6.022 x 10^23 atoms/mol = 3.18 x 10^-6 mol

Now, we can use the radioactive decay law to calculate the activity (decay rate) of the sample:

A = lambda * N = lambda * N_A * n

where N_A is Avogadro's number (6.022 x 10^23 mol^-1)

A = 0.0563/year * 6.022 x 10^23 mol^-1 * 3.18 x 10^-6 mol = 1.08 x 10^15 decays/year

Finally, we need to convert the activity from years to seconds:

A = 1.08 x 10^15 decays/year * (1 year / 3.15 x 10^7 s) = 3.41 x 10^9 decays/s

Therefore, the answer is B) 3.41E+9 decays/second.

Question 8

Approximately how many half-life periods must elapse if the activity of a radioactive isotope sample is to be reduced to 0.003 of the original value?
A)5
B)4
C)9
D)16
E)​8

Accepted Answer

To solve this problem, we need to use the half-life formula:

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

where A is the final activity, A0 is the initial activity, t is the time elapsed, and h is the half-life of the isotope.

We can rearrange this formula:

t/h = log( A / A0 ) / log(1/2)

We know that the final activity (A) is 0.003 times the initial activity (A0):

A = 0.003 * A0

So, we can substitute:

t/h = log( 0.003 ) / log(1/2)

t/h = 9.9658

We need to round this up to the nearest integer, which is 10. Therefore, approximately 10 half-life periods must elapse if the activity of a radioactive isotope sample is to be reduced to 0.003 of the original value.

The closest answer choice is C: 9.

Question 9

Over the course of 2.0 hours,44% of a radioactive material decays.What is its half-life?
A)3.45 hrs
B)2.39 hrs
C)1.12 hrs
D)0.88 hrs
E)​1.69 hrs

Accepted Answer

The answer of Over the course of 2.0 hours,44% of...

Question 10

The binding energy of a nucleus is equal to A)the energy needed to remove one of the nucleons. B)the average energy with which any nucleon is bound in the nucleus. C)the energy needed to separate all the nucleons from each other. D)the mass of the nucleus times c².

Accepted Answer

The binding energy of a nucleus is defined as the energy required to disassemble a nucleus into its constituent protons and neutrons, i.e., to separate all the nucleons from each other.

Question 11

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.
A)alpha,beta and gamma
B)gamma,beta and alpha
C)beta,gamma and alpha
D)alpha,gamma and beta

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 12

Approximately how many radioactive atoms are present in a tritium sample with an activity of 0.20 × 10−⁶ Ci and a half-life of 12.3 years? (1 Ci = 3.7 × 10¹⁰ decays/s) A)1.3E+5 B)1.1E+2 C)3.5E-6 D)4.1E+12 E)2.9E+12

Accepted Answer

We can use the following equation to relate activity (A) and number of radioactive atoms (N) of a sample:
A = λN, where λ is the decay constant.

We can also use the fact that the half-life (t1/2) is related to the decay constant by:

t1/2 = ln(2)/λ

Rearranging the first equation, we get:

N = A/λ

Substituting the value of λ from the second equation, we get:

N = A t1/2 / ln(2)

Substituting the given values, we get:

N = (0.20 × 10^0 Ci) × (12.3 years) / ln(2) × (3.7 × 10^10 decays/s)

N = 4.1 × 10^12

Therefore, the answer is D) 4.1E+12.

Question 13

A pure sample of ²²⁶Ra contains 2.3 × 10¹³ atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 × 10³ years,what is the activity of this sample? A)1.4E+10 decays/yr B)1.0E+10 decays/yr C)3.2E+2 decays/yr D)3.7E+16 decays/yr E)6.2E+6 decays/yr

Accepted Answer

The answer of A pure sample of ²²⁶Ra contains 2.3...

Question 14

If a fossil bone is found to contain 1/32 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) A)1.83E+5 years B)2.87E+4 years C)1.15E+3 years D)5.55E+3 years E)1.79E+2 years

Accepted Answer

The fossil contains 1/32 of the original Carbon-14, which means it has gone through 5 half-lives (since 2^5 = 32). Each half-life is 5,730 years, so the age of the fossil is 5 x 5,730 = 28,650 years, which is approximately 2.87E+4 years.

Question 15

Tritium has a half-life of 12.3 years.What proportion of its original radioactivity will a sample have after 2 years?
A)0.01
B)0.89
C)0.37
D)0.16
E)​0.25

Accepted Answer

The proportion of original radioactivity remaining after a certain time can be calculated using the formula: $N = N_0 \times (1/2)^{(t/T)}$, where $N$ is the remaining quantity, $N_0$ is the initial quantity, $t$ is the elapsed time, and $T$ is the half-life. Plugging in the values for tritium ($T = 12.3$ years, $t = 2$ years), we get $N/N_0 = (1/2)^{(2/12.3)}$. This calculation results in a value close to 0.89, indicating that after 2 years, a sample of tritium retains approximately 89% of its original radioactivity.

Question 16

1 Bq = ________ Ci? A)1 B)10⁶ C)2.7 × 10^-11 D)3.7 × 10¹⁰

Accepted Answer

1 Bq is equal to 2.7 × 10^-11 Ci (curie).

Question 17

A radioactive material initially is observed to have an activity of 1 600 decays/s.If six hours later it is observed to have an activity of 400 decays/s,what is its half-life?
A)six hours
B)three hours
C)twelve hours
D)seven hours
E)​not enough information

Accepted Answer

The half-life of a radioactive material is defined as the time it takes for half of the substance to decay. We can use the formula:

A = A₀(1/2)^(t/T)

where A is the final activity, A₀ is the initial activity, t is the time that has passed, and T is the half-life.

Plugging in the values given in the problem, we get:

400 = 1600(1/2)^(6/T)

Simplifying, we can divide both sides by 1600 to get:

1/4 = (1/2)^(6/T)

Taking the logarithm of both sides, we get:

log(1/4) = log[(1/2)^(6/T)]

Using logarithm rules, we can bring the exponent down:

log(1/4) = (6/T)log(1/2)

Simplifying, we get:

log(1/4) = (-6/T)

Using a calculator, we can solve for T:

T = -6/log(1/4) = 3.0 hours

Therefore, the half-life of the radioactive material is 3.0 hours, which corresponds to answer choice B.

Question 18

A pure sample of ²²⁶Ra contains 1.5 × 10¹³ 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) A)8.0E-9 Ci B)2.1E+2 Ci C)5.6E-9 Ci D)1.8E-1 Ci E)2.5E-1 Ci

Accepted Answer

The decay rate can be calculated using the formula:

Decay rate = (ln 2 / half-life) x number of atoms

Decay rate = (ln 2 / 1.6 × 10^13 years) x (1.5 × 10^13 atoms)
Decay rate = 0.43 x 1.5
Decay rate = 0.65

Converting the decay rate to Ci:

1 Ci = 3.7 × 10^10 decays/s
0.65 decays/s = 0.65 / 3.7 × 10^10 = 1.76 × 10^-11 Ci

Therefore, the best choice is C) 5.6E-9 Ci, which is closest to the calculated value of 1.76 × 10^-11 Ci (rounded to two significant figures).

Question 19

The fact that the binding energy per nucleon does not depend very strongly on the volume of the nucleus indicates that
A)the strong nuclear force saturates.
B)nucleons don't move throughout the nucleus.
C)all nuclei have the same volume.
D)the radius of a nucleus is directly proportional to the number of nucleons.

Accepted Answer

The fact that the binding energy per nucleon does not depend very strongly on the volume of the nucleus indicates that there is a saturation effect in the strong nuclear force. As more nucleons are added to the nucleus, the attractive force between nucleons becomes saturated, leading to a nearly constant binding energy per nucleon.

Question 20

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) A)2.24 MeV/nucleon B)2.45 MeV/nucleon C)2.66 MeV/nucleon D)2.86 MeV/nucleon

Accepted Answer

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

Quiz 29: Nuclear Physics

Tritium has a half-life of 12.3 years.How many years will elapse when the radioactivity of a tritium sample diminishes to 35% of its original value?

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

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)

The half-life of radioactive Technetium-99 is 6.0 hours.Find the number of 99Tc nuclei necessary to produce a sample of activity 1.9 µCi.(1 Ci = 3.7 × 1010 decays/second)

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 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 1.91 × 1018 tritium atoms,what is its activity in decays/second? (1 year = 3.15 × 107 s)

Approximately how many half-life periods must elapse if the activity of a radioactive isotope sample is to be reduced to 0.003 of the original value?

Over the course of 2.0 hours,44% of a radioactive material decays.What is its half-life?

The binding energy of a nucleus is equal to

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.

Approximately how many radioactive atoms are present in a tritium sample with an activity of 0.20 × 10−6 Ci and a half-life of 12.3 years? (1 Ci = 3.7 × 1010 decays/s)

A pure sample of 226Ra contains 2.3 × 1013 atoms of the isotope.If the half-life of 226Ra = 1.6 × 103 years,what is the activity of this sample?

If a fossil bone is found to contain 1/32 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)

Tritium has a half-life of 12.3 years.What proportion of its original radioactivity will a sample have after 2 years?

1 Bq = ________ Ci?

A radioactive material initially is observed to have an activity of 1 600 decays/s.If six hours later it is observed to have an activity of 400 decays/s,what is its half-life?

A pure sample of 226Ra contains 1.5 × 1013 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)

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

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

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)

The half-life of radioactive Technetium-99 is 6.0 hours.Find the number of ⁹⁹Tc nuclei necessary to produce a sample of activity 1.9 µCi.(1 Ci = 3.7 × 10¹⁰ decays/second)

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 1.91 × 10¹⁸ tritium atoms,what is its activity in decays/second? (1 year = 3.15 × 10⁷ s)

Approximately how many radioactive atoms are present in a tritium sample with an activity of 0.20 × 10−⁶ Ci and a half-life of 12.3 years? (1 Ci = 3.7 × 10¹⁰ decays/s)

A pure sample of ²²⁶Ra contains 2.3 × 10¹³ atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 × 10³ years,what is the activity of this sample?

If a fossil bone is found to contain 1/32 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)

A pure sample of ²²⁶Ra contains 1.5 × 10¹³ 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)

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)