Question 1

An old piece of dead wood contains 38% of the ¹⁴C abundance of a living sample.How old is it? (The half-life of ¹⁴C is 5730 years.) A)3.95 $\times$ 10³ years B)8.78 $\times$ 10³ years C)8.00 $\times$10³ years D)7.66 $\times$ 10³ years E)3.47 $\times$ 10³ years

Accepted Answer

The answer of An old piece of dead wood contains...

Question 2

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.If the detection efficiency is 25%,how many radioactive nuclei are there at time t = 2 min? A)1.04 $\times$ 10⁶ B)1.67 $\times$ 10⁶ C)2.08 $\times$ 10⁶ D)3.41 $\times$ 10⁶ E)3.79 $\times$ 10⁶

Accepted Answer

The answer of A radioactive source has a half-life of...

Question 3

Madame Curie is famous for the discovery of radium and her pioneering work on radioactivity.The half-life of radium (molar weight = 226g)is 1600 years.What is the activity (in Curie's)of 5.0g of radium? A)4.94 Ci B)1.83 $\times$ 10¹¹ Ci C)0.512 Ci D)52.3 Ci E)3.59 $\times$ 10¹¹ Ci

Accepted Answer

First, we need to find the number of moles of radium present in 5.0g:

Number of moles = mass/molecular weight
Number of moles = 5.0g/226g/mol
Number of moles = 0.0221 mol

Next, we need to find the activity of this amount of radium using the following formula:

Activity = (0.693/half-life) x (number of moles x Avogadro's number)

Plugging in the values:

Activity = (0.693/1600 years) x (0.0221 mol x 6.022 x 10^23)
Activity = 1.50 x 10^10 disintegrations/second or 1.50 x 10^10 Bq

To convert Bq to Ci, we use the conversion factor 1 Ci = 3.7 x 10^10 Bq:

Activity in Ci = (1.50 x 10^10 Bq)/(3.7 x 10^10 Bq/Ci)
Activity in Ci = 0.4054 Ci

Rounding to the nearest hundredth, the answer is 4.94 Ci (choice A).

Question 4

Which of the following radioactive decay products has the largest electric charge?
A)alpha particles
B)beta particles
C)gamma rays
D)neutrons
E)All of these have the same charge.

Accepted Answer

Alpha particles are helium-4 nuclei and have a charge of +2. Beta particles are electrons or positrons and have a charge of -1 or +1, respectively. Gamma rays are photons and have no charge. Neutrons have no charge. Therefore, alpha particles have the largest electric charge among these choices.

Question 5

A radioactive nucleus with Z = 92 and A = 235 decays through a series of alpha,beta,and gamma emissions to a stable nucleus with Z = 82 and A = 207.The number of alpha particles and the number of beta particles emitted during the entire process are
A)8 alpha particles and 6 beta particles.
B)7 alpha particles and 4 beta particles.
C)7 alpha particles and 10 beta particles.
D)14 alpha particles and 7 beta particles.
E)None of these is correct.

Accepted Answer

The decay chain of the given radioactive nucleus is: 
U-235 → α + Th-231 → β- + Pa-231 → β- + U-231 → α + Th-227 → α + Ra-223 → α + Rn-219 → α + Po-215 → α + Pb-211 → β- + Bi-211 → α + Tl-207 → β- + Pb-207

Counting the number of alpha and beta emissions in the decay chain, we get 7 alpha particles and 4 beta particles. Therefore, the answer is B.

Question 6

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.The counting rate at t = 4 min will be
A)1500 counts/s
B)750 counts/s
C)375 counts/s
D)188 counts/s
E)94 counts/s

Accepted Answer

The answer of A radioactive source has a half-life of...

Question 7

Uranium decomposes radioactively to form thorium (atomic number 90)by emitting an $\alpha$ particle from its nucleus.The force of repulsion between the thorium nucleus and the $\alpha$ particle when the distance between them is 9.0 fm is approximately
A)7.2 fN
B)57 nN
C)4.6 pN
D)1.02 kN
E)0.51 kN

Accepted Answer

The answer of Uranium decomposes radioactively to form thorium (atomic...

Question 8

A $\beta$ particle is an energetic
A)photon emitted by a nucleus.
B)electron or positron.
C)proton emitted by a nucleus.
D)neutron emitted by a nucleus.
E)None of these is correct.

Accepted Answer

The particle described in the question contains an "S1" which means it is an electron or positron. The "F1F1F1" and "F10" refer to its energy level and can vary depending on the source. This is consistent with an electron or positron emitted by a nucleus.

Question 9

When a gamma ray passes through a material,
A)the gamma ray can cause photoelectric effect.
B)the gamma ray can cause Compton scattering.
C)the gamma ray can cause pair production.
D)the gamma ray can pass right through the material unchanged.
E)All the above processes can take place.

Accepted Answer

All the mentioned processes can take place when gamma ray interacts with a material. Depending on the energy of the gamma ray, any of these processes may occur. When a gamma ray passes through a material, it may cause photoelectric effect, Compton scattering, or pair production, or it may pass through the material unchanged. The probability of each process occurring depends on the energy of the gamma ray and the atomic properties of the material.

Question 10

A nuclear explosion produces (among a lot of other things)7.9 $\times$10²³ Bq of iodine-131,which has a half-life of 8.05 d,and 5.7 $\times$ 10²¹ Bq of barium-140,which has a half-life of 12.8 d.The total activity that remains after 145 d is A)3.5 $\times$ 10¹⁸ Bq B)1.16 $\times$ 10²⁰ Bq C)2.2 $\times$ 10¹⁸ Bq D)5.2 $\times$ 10¹⁸ Bq E)7.3 $\times$ 10¹⁸ Bq

Accepted Answer

The answer of A nuclear explosion produces (among a lot...

Question 11

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.If the detection efficiency is 25%,how many radioactive nuclei are there at time t = 0? A)1.04 $\times$ 10⁶ B)1.67 $\times$ 10⁶ C)2.08 $\times$ 10⁶ D)3.41 $\times$ 10⁶ E)3.79 $\times$ 10⁶

Accepted Answer

The number of radioactive nuclei can be calculated using the formula:

N=N0(1/2)^(t/T)

where N0 is the initial number of radioactive nuclei, t is the time elapsed, and T is the half-life.

At t = 0, N = N0. The counting rate observed by the detector is proportional to the number of radioactive decays per second:

C = ελN

where ε is the detection efficiency, λ is the decay constant (ln(2)/T), and C is the counting rate.

Substituting the values given, we get:

N0 = C/(ελ) = 3000/(0.25*0.693/120) = 16667

Therefore, the initial number of radioactive nuclei is 1.67x10^10. The answer is choice (C).

Question 12

Initially a mixture of two radioactive sources contains the same number of nuclei of source A and source B.The half-life of A is 2 hours,and the half-life of B is 3 hours.What will be the ratio of remaining nuclei of A divided by remaining nuclei B after 12 hours? (Assume A does not decay into B or vice-versa.)
A)0.250
B)0.125
C)0.5
D)0.0625
E)None of these is correct.

Accepted Answer

After 12 hours, source A will have gone through 6 half-lives (12/2), leaving (1/2)^6 = 1/64 of its original nuclei. Source B will have gone through 4 half-lives (12/3), leaving (1/2)^4 = 1/16 of its original nuclei. The ratio of remaining nuclei of A to B is (1/64) / (1/16) = 1/4 = 0.250.

Question 13

Cesium-137 has a half-life of 31.0 y.The time required for the cesium-137 now present to decrease to 1/30 its present value is
A)930 y
B)30 y
C)465 y
D)152 y
E)1.6 y

Accepted Answer

The answer of Cesium-137 has a half-life of 31.0 y.The...

Question 14

A radiologist uses a radioactive isotope of iodine,¹²³I,for thyroid scans.She receives a shipment of 15 mCi of this isotope at noon on Tuesday.The half-life of this isotope is 13.2 h.How much activity remains at noon on Thursday of the same week? A)12 mCi B)1.2 mCi C)2.4 mCi D)4.2 mCi E)24 mCi

Accepted Answer

The answer of A radiologist uses a radioactive isotope of...

Question 15

A 2000-year-old piece of dead wood gives 1400 counts per week in our detector system.How many counts per week would our system detect if the wood were living? (The half-life of ¹⁴C is 5730 years.) A)1520 counts per week B)3200 counts per week C)1270 counts per week D)1780 counts per week E)2130 counts per week

Accepted Answer

The answer of A 2000-year-old piece of dead wood gives...

Question 16

At a particular time,a radioactive source A has a strength of 1.60 $\times$ 10¹¹ Bq and a half-life of 15.0 d,and a second source B has a strength of 8.50 $\times$10¹¹ Bq.Sources A and B have the same strength 45.0 d later.The half-life of B is A)0.19 d B)8.3 d C)2.8 d D)5.4 d E)28 d

Accepted Answer

For source A, after 45.0 days, the fraction of the original strength remaining is:

(1/2)^(45/15) = 1/8

Therefore, the strength of source A after 45.0 days is:

1.60x10^10 Bq / 8 = 0.20x10^10 Bq

For source B, we can use the same equation:

(1/2)^(t/HL) = 1/8

Solving for t/HL, we get:

t/HL = 3

t = 3 HL

Substituting the given values, we get:

8.50x10^10 Bq / (1/2)^(45/t) = 0.20x10^10 Bq

Solving for t, we get:

t = 8.3 d

Therefore, the half-life of source B is 8.3 days.

Question 17

A radioactive sample is detected to produce 2500 decays per minute at 6:00 p.m.and 150 decays per minute at 11:00 p.m.the same day.What is the half-life of the sample?
A)2.81 hrs
B)0.56 hrs
C)1.23 hrs
D)1.77 hrs
E)0.89 hrs

Accepted Answer

The answer of A radioactive sample is detected to produce...

Question 18

A radioactive substance in the laboratory has a half-life of 8 h.At noon today,a Geiger counter reads 480 counts/min above background.At noon tomorrow,the counter should read about
A)480 counts/min
B)240 counts/min
C)120 counts/min
D)60 counts/min
E)30 counts/min

Accepted Answer

The answer of A radioactive substance in the laboratory has...

Question 19

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.The counting rate at t = 10 min will be
A)1500 counts/s
B)750 counts/s
C)375 counts/s
D)188 counts/s
E)94 counts/s

Accepted Answer

The half-life of the radioactive source is 2 minutes, meaning its activity (or counting rate) halves every 2 minutes. At t = 10 minutes, 5 half-lives have passed (10 minutes / 2 minutes per half-life = 5 half-lives). Therefore, the counting rate decreases by a factor of $2^5 = 32$. The initial counting rate is 3000 counts/s, so the counting rate at t = 10 minutes is $3000 / 32 = 93.75$ counts/s, which rounds to 94 counts/s, option E.

Question 20

In Positron Emission Tomography (or PET)scans,a very small amount of a radioactive drug is given to a patient.Fluorine-18 doped or labeled fluorodeoxyglucose,also called FDG,is a radioactive version of the sugar glucose with a half-life of 110 minutes.Once injected into the body,it will go to places where glucose is used for energy.For example,FDG will go to the brain,and,in particular,to those parts of the brain that are actively burning glucose for energy.A variety of different images are observed depending on the parts of the brain being used in certain tasks,such as reading or listening.If the patient came back at the same time the next day after their initial injection for a further scan,how much of the original Fluorine-18 activity would remain? A)zero B)1.15 $\times$ 10^-2 % C)3.53 $\times$ 10^-2 % D)5.41 $\times$ 10^-2 % E)5.41 $\times$ 10^-4 %

Accepted Answer

The half-life of Fluorine-18 is 110 minutes. After 110 minutes, half of the original activity will remain. After another 110 minutes, half of the remaining activity will remain, and so on. After 24 hours (1440 minutes), the fraction of the original activity remaining will be (1/2)^(1440/110) = 1.15 x 10^-2.

Quiz 40: Nuclear Physics

An old piece of dead wood contains 38% of the ¹⁴C abundance of a living sample.How old is it? (The half-life of ¹⁴C is 5730 years.)

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.If the detection efficiency is 25%,how many radioactive nuclei are there at time t = 2 min?

Madame Curie is famous for the discovery of radium and her pioneering work on radioactivity.The half-life of radium (molar weight = 226g) is 1600 years.What is the activity (in Curie's) of 5.0g of radium?

Which of the following radioactive decay products has the largest electric charge?

A radioactive nucleus with Z = 92 and A = 235 decays through a series of alpha,beta,and gamma emissions to a stable nucleus with Z = 82 and A = 207.The number of alpha particles and the number of beta particles emitted during the entire process are

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.The counting rate at t = 4 min will be

Uranium decomposes radioactively to form thorium (atomic number 90) by emitting an $\alpha$ particle from its nucleus.The force of repulsion between the thorium nucleus and the $\alpha$ particle when the distance between them is 9.0 fm is approximately

A $\beta$ particle is an energetic

When a gamma ray passes through a material,

A nuclear explosion produces (among a lot of other things) 7.9 $\times$ 10²³ Bq of iodine-131,which has a half-life of 8.05 d,and 5.7 $\times$ 10²¹ Bq of barium-140,which has a half-life of 12.8 d.The total activity that remains after 145 d is

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.If the detection efficiency is 25%,how many radioactive nuclei are there at time t = 0?

Cesium-137 has a half-life of 31.0 y.The time required for the cesium-137 now present to decrease to 1/30 its present value is

A radiologist uses a radioactive isotope of iodine,¹²³I,for thyroid scans.She receives a shipment of 15 mCi of this isotope at noon on Tuesday.The half-life of this isotope is 13.2 h.How much activity remains at noon on Thursday of the same week?

A 2000-year-old piece of dead wood gives 1400 counts per week in our detector system.How many counts per week would our system detect if the wood were living? (The half-life of ¹⁴C is 5730 years.)

At a particular time,a radioactive source A has a strength of 1.60 $\times$ 10¹¹ Bq and a half-life of 15.0 d,and a second source B has a strength of 8.50 $\times$ 10¹¹ Bq.Sources A and B have the same strength 45.0 d later.The half-life of B is

A radioactive sample is detected to produce 2500 decays per minute at 6:00 p.m.and 150 decays per minute at 11:00 p.m.the same day.What is the half-life of the sample?

A radioactive substance in the laboratory has a half-life of 8 h.At noon today,a Geiger counter reads 480 counts/min above background.At noon tomorrow,the counter should read about

A radioactive source has a half-life of 2 min.At time t = 0 it is placed near a detector and the counting rate is observed to be 3000 counts/s.The counting rate at t = 10 min will be