Question 1

Nuclear reactions: In the nuclear reaction n + U → X + 2e^- n is a neutron and e^- is an electron, and the neutrinos have not been shown. Determine the atomic mass and atomic number of the missing nuclear product X, and write X in the standard form. It is NOT necessary to identify which atom X is.

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n +...

Question 2

Radioactive dating: An ancient rock is found to contain ⁴⁰Ar gas, indicating that of the ⁴⁰K in the rock has decayed since the rock solidified. Any argon would have boiled out of liquid rock. The half-life of ⁴⁰K is 1.25 billion years. How long ago did the rock solidify?

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Question 3

Nuclear fission: How much energy is released in the total fission of of The average energy per fission is 200.0 MeV. (1 u = 931.5 MeV/c² = 1.6605 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J)

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Question 4

Radioactive dating: Living matter has 1.3 × 10^-10 % of its carbon in the form of ¹⁴C which has a half-life of 5730 y. A mammoth bone has a 300-g sample of carbon separated from it, and the sample is found to have an activity of 20 decays per second. How old is the bone?

Accepted Answer

The age of the bone can be determined using the decay formula for radioactive isotopes, which relates the current activity to the initial activity, taking into account the half-life of the isotope. Given the half-life of C-14 and the current activity, calculations will show that the bone is approximately 10,900 years old.

Question 5

Nuclear reactions: For the missing product X in the reaction
neutron +  U →  Ba + X + 3 neutrons
determine the atomic mass and atomic number of X, and write X in the standard form. It is NOT necessary to identify which atom X is.

Accepted Answer

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Question 6

Nuclear fission: In the fission reaction  U + neutron →  Ba +  Kr + x neutrons, what is the number x of neutrons produced?

Accepted Answer

In the given fission reaction, one neutron is absorbed by uranium-235 nucleus and it splits into two lighter nuclei, barium-144 and krypton-89, and three more neutrons are released. Hence, the number of neutrons produced is 3. Therefore, the correct choice is D) 3.

Question 7

Nuclear fission: If a 2.0-MeV neutron released in a fission reaction loses half of its energy in each moderator collision, how many collisions are needed to reduce its energy to (1/25) eV?

Accepted Answer

The energy of the neutron is halved with each collision. Starting from 2.0 MeV, we need to find how many halvings are required to reach (1/25) eV. Convert 2.0 MeV to eV: 2.0 MeV = 2,000,000 eV. The energy after n collisions is (2,000,000 eV) * (1/2)^n. Set this equal to (1/25) eV and solve for n: (1/2)^n = (1/25) / 2,000,000. Solving gives n ≈ 26.

Question 8

Nuclear fission: An excited U* nucleus undergoes fission into two fragments, as shown: U* → Ba + Kr The following atomic masses are known: Kr: 91.926270 u Ba: 143.922845 u U*: 236.045563 u What is the reaction energy, in MeV, for this process? (1 u = 1.6605 × 10^-27 kg = 931.5 MeV/c²)

Accepted Answer

The answer of Nuclear fission: An excited  U* nucleus...

Question 9

Reaction energy: Find the reaction energy (Q value) of the reaction N + He → O + H, given the following masses: N: 14.003074 u He: 4.002603 u O : 16.999131 u H: 1.007825 u (1 u = 931.5 MeV/c²)

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Question 10

Nuclear reactions: For the missing product X in the nuclear reaction
neutron +  O → X + alpha particle
determine the atomic mass and atomic number of X, and write X in the standard form. It is NOT necessary to identify which atom X is.

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Question 11

Reaction energy: The reaction energy (Q value) for a particular reaction is -2.4 MeV, and the reaction's threshold energy is 9.60 MeV. What is the ratio of the mass of the incident particle to the mass of the stationary target nucleus?

Accepted Answer

The Q value relates to the difference in mass between the initial and final states of a nuclear reaction. We can use the following formula to relate the Q value to the masses of the particles involved:

Q = (M_initial - M_final)c^2

where c is the speed of light. Let m be the mass of the incident particle and M be the mass of the stationary target nucleus. Then the masses of the particles in the initial and final states are:

Initial state: m + M
Final state: (m + M - Δm) + ΔM

where Δm and ΔM are the changes in mass due to the reaction. We can simplify these expressions by assuming that the incident particle is highly energetic and relativistic, so that its energy is much greater than its rest mass. In this case, we can neglect the mass of the incident particle in the initial state, and assume that in the final state the target nucleus is left in a highly excited state and its mass is unchanged. Then the expressions become:

Initial state: M
Final state: M - Δm + ΔM

The Q value is given as -2.4 MeV, which is the energy released in the reaction. To relate this to the masses of the particles, we use the formula:

Q = (M - (M - Δm + ΔM))c^2
-2.4 MeV = Δmc^2 - ΔMc^2

The threshold energy is given as 9.60 MeV, which is the minimum energy needed for the reaction to occur. This means that the incident particle has kinetic energy equal to the threshold energy, which can be related to its velocity and mass through the formula:

K = (1/2)mv^2 = (γ - 1)mc^2
9.60 MeV = (γ - 1)mc^2
where γ is the Lorentz factor, given by γ = (1 - v^2/c^2)^(-1/2).

We can eliminate the velocity v by using the formula for the relativistic energy E = γmc^2:

E = K + mc^2
9.60 MeV = (γ - 1 + 1)mc^2
γ = 10.6

Substituting this value of γ into the formula for the threshold energy, we get:

9.60 MeV = (γ - 1)mc^2
9.60 MeV = 9.6mc^2
mc^2 = 1 MeV

Substituting this value of mc^2 into the formula for the Q value, we get:

-2.4 MeV = Δm - ΔM

Since the reaction involves the absorption of an incident particle, we can assume that ΔM is positive (the target nucleus gains mass), which means that Δm is negative (the incident particle loses mass). Then we have:

-2.4 MeV = -|Δm| + ΔM
|Δm| = 2.4 MeV - ΔM

The ratio of the mass of the incident particle to the mass of the target nucleus is given by:

m/M = (γ^2 - 1)^(-1/2)
m/M = (10.6^2 - 1)^(-1/2)
m/M = 0.327

Substituting this value into the expression for |Δm|, we get:

|Δm| = 0.327ΔM

Substituting this expression into the equation for the Q value, we get:

-2.4 MeV = -0.327ΔMc^2 + ΔMc^2
ΔM = 7.35 MeV/c^2

Substituting this value of ΔM into the expression for the mass ratio, we get:

m/M = 0.327
m = 0.327M
Δm = -2.4 MeV + ΔM = 5.95 MeV/c^2

The correct answer is therefore (C) 3.0.

Question 12

Nuclear reactions: In the nuclear reaction  B +  He →  H + X, which of the following is the missing nuclear product X?

Accepted Answer

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Question 13

Reaction energy: The detonation of a certain nuclear device results in a mass decrease of between the initial and the final ingredients. How much energy is released by this detonation? (c = 3.00 × 10⁸ m/s)

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Question 14

Nuclear reactions: A proton strikes an  O nucleus producing  F and another particle. What is the other particle?

Accepted Answer

The proton must combine with 10 neutrons to create the fluorine isotope, which means it must have produced a neutron in the process. Therefore, the best choice for the other particle is a neutron.

Question 15

Nuclear fission: When a neutron (n) collides with a uranium-235 nucleus it can induce a variety of fission reactions. One such reaction is U + n → Xe + Sr + 2n. How much energy is released in this reaction, given the following mass values: Xe: 139.921620 u Sr : 93.915367 u U: 235.043924 u n: 1.008665 u (1 u = 931.494 MeV/c²)

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Question 16

Radioactive dating: Today, the uranium found on Earth contains 0.720% ²³⁵U (with a half-life of 0.700 billion years) and 99.28% ²³⁸U (with a half-life of 4.50 billion years). At a time 2.20 billion years ago, what percent of the uranium on Earth was ²³⁸U (assuming that no other uranium isotopes were present)?

Accepted Answer

To find the percentage of Uranium-238 (U-238) 2.20 billion years ago, we use the decay formula for half-life. The half-life of U-238 is 4.50 billion years. The formula to calculate the remaining amount of a substance after a certain time is: $$N = N_0 \times (1/2)^{t/T}$$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life of the substance. Since we are interested in the percentage of U-238, we can simplify our calculations by focusing on the ratio of U-238 to the total uranium, which initially is 99.28%. The time elapsed is 2.20 billion years, and the half-life of U-238 is 4.50 billion years. However, since the question asks for the percentage of U-238 2.20 billion years ago, and given that U-238 decays much slower than U-235, the percentage of U-238 would have been slightly higher than it is today, but the calculation provided does not directly apply to finding the initial percentage of U-238. Instead, we should consider the decay of U-235 and its effect on the overall uranium composition. The rapid decay of U-235 compared to U-238 would mean that U-238's relative abundance would increase over time. Given the options and the nature of radioactive decay, where U-238 becomes more dominant over long periods due to its longer half-life, the correct answer reflects the need to understand the dynamics of both isotopes' decay rates and their impact on the uranium composition over time. The calculation mistake here is in the explanation approach, which should focus on the comparative decay of both isotopes and the fact that the question implies a calculation error by not directly applying the decay formula to find the initial percentage of U-238. The correct approach would involve understanding that U-238, with its longer half-life, would have constituted a larger percentage of the uranium on Earth 2.20 billion years ago due to the faster decay of U-235, leading to an increase in the relative abundance of U-238 over time.

Question 17

Radioactive dating: An archaeologist finds the ¹⁴C in a sample of of material to be decaying at 107 counts per second. A modern 1.00-g sample of the same material decays at 151 counts per second. The half-life of ¹⁴C is 5730 y. How old is the sample?

Accepted Answer

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Question 18

Nuclear fission: An excited U^* nucleus undergoes fission into two fragments, as shown: U* → Ba + Kr The following atomic masses are known: Kr: 91.926270 u Ba: 143.922845 u U*: 236.045563 u Assume, at a given instant, that the two fission fragments are spherical, just barely in contact, and carry spherically symmetric charge distributions. At that instant, what is the electrostatic interaction energy of the two fragments, in MeV? (1 u = 1.6605 × 10^-27 kg = 931.5 MeV/c², 1/4πε₀ = 9.0 × 10⁹ N * m²/C²)

Accepted Answer

The answer of Nuclear fission: An excited U^* nucleus...

Question 19

Nuclear fusion: Calculate the amount of energy that is released in the fusion reaction ²H + ²H → ⁴He, given the masses: ²H: 2.014102 u ⁴He: 4.002603 u (1 u = 931.5 MeV/c²)

Accepted Answer

The total mass of the reactants is 2(2.014102 u) = 4.028204 u. The mass of the product is 4.002603 u. The mass difference is 4.028204 u - 4.002603 u = 0.025601 u. Converting this mass difference into energy using E = mc^2 and given 1 u = 931.5 MeV/c^2, the energy released is 0.025601 u * 931.5 MeV/u = 23.84 MeV, which rounds to approximately 24 MeV.

Question 20

Reaction energy: Calculate the reaction energy (Q value) for the reaction 7Li + 1H → 4He + 4He, given the following masses: 7Li: 7.016005 u 1H: 1.007825 u 4He: 4.002603 u (1 u = 931.5 MeV/c²)

Accepted Answer

The reaction equation can be written as follows:

7Li + 1H → 2(4He)

Using the given masses,

mass of the left-hand side = (7.016005 + 1.007825) u = 8.02383 u.

mass of the right-hand side = 2(4.002603) u = 8.005206 u.

The difference between the masses is Δm = (mass of LHS – mass of RHS) = 8.02383 u – 8.005206 u = 0.018624 u.

The reaction energy (Q value) can be calculated using Einstein's famous equation E = mc², where E is the energy, m is the mass difference, and c is the speed of light.

Q = Δmc²

= (0.018624 u) x (931.5 MeV/c²)

= 17.35 MeV (rounded to two decimal places).

Therefore, the best choice is D with the explanation as provided above.

Quiz 34: Geometric Optics

Nuclear reactions: In the nuclear reaction n + U → X + 2e- n is a neutron and e- is an electron, and the neutrinos have not been shown. Determine the atomic mass and atomic number of the missing nuclear product X, and write X in the standard form. It is NOT necessary to identify which atom X is.

Radioactive dating: An ancient rock is found to contain 40Ar gas, indicating that of the 40K in the rock has decayed since the rock solidified. Any argon would have boiled out of liquid rock. The half-life of 40K is 1.25 billion years. How long ago did the rock solidify?

Nuclear fission: How much energy is released in the total fission of of The average energy per fission is 200.0 MeV. (1 u = 931.5 MeV/c2 = 1.6605 × 10-27 kg, 1 eV = 1.60 × 10-19 J)

Radioactive dating: Living matter has 1.3 × 10-10 % of its carbon in the form of 14C which has a half-life of 5730 y. A mammoth bone has a 300-g sample of carbon separated from it, and the sample is found to have an activity of 20 decays per second. How old is the bone?

Nuclear reactions: For the missing product X in the reaction neutron + U → Ba + X + 3 neutrons determine the atomic mass and atomic number of X, and write X in the standard form. It is NOT necessary to identify which atom X is.

Nuclear fission: In the fission reaction U + neutron → Ba + Kr + x neutrons, what is the number x of neutrons produced?

Nuclear fission: If a 2.0-MeV neutron released in a fission reaction loses half of its energy in each moderator collision, how many collisions are needed to reduce its energy to (1/25) eV?

Nuclear fission: An excited U* nucleus undergoes fission into two fragments, as shown: U* → Ba + Kr The following atomic masses are known: Kr: 91.926270 u Ba: 143.922845 u U*: 236.045563 u What is the reaction energy, in MeV, for this process? (1 u = 1.6605 × 10-27 kg = 931.5 MeV/c2)

Reaction energy: Find the reaction energy (Q value) of the reaction N + He → O + H, given the following masses: N: 14.003074 u He: 4.002603 u O : 16.999131 u H: 1.007825 u (1 u = 931.5 MeV/c2)

Nuclear reactions: For the missing product X in the nuclear reaction neutron + O → X + alpha particle determine the atomic mass and atomic number of X, and write X in the standard form. It is NOT necessary to identify which atom X is.

Reaction energy: The reaction energy (Q value) for a particular reaction is -2.4 MeV, and the reaction's threshold energy is 9.60 MeV. What is the ratio of the mass of the incident particle to the mass of the stationary target nucleus?

Nuclear reactions: In the nuclear reaction B + He → H + X, which of the following is the missing nuclear product X?

Reaction energy: The detonation of a certain nuclear device results in a mass decrease of between the initial and the final ingredients. How much energy is released by this detonation? (c = 3.00 × 108 m/s)

Nuclear reactions: A proton strikes an O nucleus producing F and another particle. What is the other particle?

Radioactive dating: An archaeologist finds the 14C in a sample of of material to be decaying at 107 counts per second. A modern 1.00-g sample of the same material decays at 151 counts per second. The half-life of 14C is 5730 y. How old is the sample?

Nuclear fusion: Calculate the amount of energy that is released in the fusion reaction 2H + 2H → 4He, given the masses: 2H: 2.014102 u 4He: 4.002603 u (1 u = 931.5 MeV/c2)

Reaction energy: Calculate the reaction energy (Q value) for the reaction 7Li + 1H → 4He + 4He, given the following masses: 7Li: 7.016005 u 1H: 1.007825 u 4He: 4.002603 u (1 u = 931.5 MeV/c2)