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Practice Problems for Final

Practice problems for the final are listed below. This exam will cover all material from the course

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Topic: Radioactivity

  1. Determine decay products from a spontaneous nuclear decay

    Complete each of the following nuclear reactions.
    • 212Po -> 4He + ... + Q
    • 14C -> ... + 14N + anti-ν + Q
    • 59Fe -> b- + ... + anti-ν + Q
    • 145Tm -> 1H + ... + Q
    • 22Na -> ... + 22Ne + ν + Q
    • 122Rh -> 121Rh + ...
    Answer
    • 208Pb
    • b-
    • 59Co
    • 144Er
    • b+
    • 0n
    Explanation
    • The emission of an alpha particle from 212Po will reduce the number of protons by 2 and number of neutrons by 2, producing 208Pb
    • The nuclear mass remains constant, while the proton number is increased by 1 and the neutron number decreased by 1. This process is b- decay.
    • The process of b- decay increases the proton number by 1 and decreases the neutron number by 1. The resulting nucleus is 59Co
    • Emission of a proton from 145Tm will reduce the proton number and total mass by one unit each. The final nucleus is therefore 144Er.
    • The nuclear mass remains constant, while the proton number is decreased by 1 and the neutron number increased by 1. This process is positron, or b+ decay.
    • Emission of a neutron from 122Rh will reduce the neutron number and total mass by one unit each. The final nucleus is therefore 121Rh.

  2. Radioactive Decay Modes

    Consider the following statements pertaining to the different types of radioactive decay. Determine whether each statement is true or false.
    • Beta radiation consists of streams of high-energy protons emitted by an unstable nucleus.
    • A positron is a particle with the same mass as an electron, but opposite charge.
    • Gamma radiation consist of high-energy electromagnetic radiation.
    • Emission of an alpha particle reduces the nuclear charge by 2 units, and the nuclear mass by 3 units.
    • Electron capture is a process that competes with spontaneous emission of fast, negatively-charged electrons.
    Answer
    • False
    • True
    • True
    • False
    • False
    Explanation
    • Beta radiation consists of high-energy electrons emitted by an unstable nucleus.
    • There are two types of beta radiation: fast, negative electron emission and fast, positive electron emission. The positron is the later, and considered the anti-particle of the electron.
    • Gamma radiation consists of photons, that are quanitized units of electromagnetic radiation with very short wavelength.
    • An alpha particle is a 4He nucleus, and emission of an alpha will reduce the nuclear charge by 2 units, and the nuclear mass by 4 units.
    • The capture of an ortibal electron, leading effectively to the conversion of a proton to a neutron inside the atomic nucleus, is a process that competes with positron emission. Positron emission can only occur if the energy difference between initial and final states is greater than 1.02 MeV.

  3. Trends in Radioactive Decay Across the Nuclear Chart

    Consider the following statements regarding the trends in radioactive decay across the chart of the nuclides. Determine whether each statement is true or false.
    • Heavy nuclei tend to decay by alpha emission.
    • Nuclei with higher neutron-to-proton ratios than those along the line of stability decay by positron emission
    • Electron capture is a possible decay pathway for nuclei with lower neutron-to-proton ratios than those along the line of stability decay by positron emission
    • Nuclei with even numbers of both protons and neutrons compose the greatest number of stable isotopes.
    • All nuclei with proton number greater than 84 are radioactive.
    Answer
    • True
    • False
    • True
    • True
    • True
    Explanation
    • Alpha decay is the predominant decay mode for heavy nuclei above A = 150.
    • Nuclei with more abundant neutrons than those at stability will decay by b- emission.
    • Electron capture and positron decay are the two primary decay modes for nuclei that have a lower neutron-to-proton ratio that their counterparts at the valley of stability.
    • Nuclei with even numbers of proton and neutrons have extra binding due to the pairing interaction, which is favored by the strong nuclear force.
    • The last "stable" nucleus is 209Bi. Some isotopes of uranium (Z = 92) have very long lifetimes (109 years) and can be found in the natural environment. But they are still radioactive, and emit alpha and beta particles over time.

Topic: Radioactive Decay Rates

  1. Quantitative Calculation of Radioactive Decay Rates

    How much time is required for a 5.75-mg sample of 51Cr to decay to 1.50 mg if it has a half-life of 27.8 days?
    Answer
    53.9 days
    Explanation
    Radioactive decay follows the first order rate law:
    ln [A]t = -kt + ln[A]o, where
    [A]t is the concentration at time t

    k = 0.693/t1/2

    [A]t is the initial concentration
    For the problem at hand, we can directly replace concentration with mass, since the MW of the sample does not change (it remains 51Cr)
    ln (1.5 mg) = -(0.693/27.8 d)*t + ln(5.75 mg)
    ln (1.5 mg) = -(0.693/27.8 d)*t + ln(5.75 mg)
    0.4055 = -0.0250*t + 1.75
    t = 53.9 days

  2. Age of Acheological Sample from Carbon Dating

    A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample that is due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5,715 years. What is the age of the archeological sample?
    Answer
    2,230 years old
    Explanation
    Radiocarbon dating works on the principle that living objects intake natural 14C radioactivity. This constant intake keeps the 14C activity constant in living things. When the object dies, it can no longer uptake 14C, and it will decay away with a half-life of 5,715 years.
    For the problem at hand, the fresh wood sample represents the constant activity of 14C expected in wood. The activity in the archeological sample is smaller because the 14C has decay away. To determine the decay time:
    ln (Nsample/Nfresh) = - (0.693/t1/2)*time
    Nsample = 11.6 dps
    Nfresh = 15.2 dps
    t1/2 = 5,175 years
    ln (11.6 dps/15.2 dps) = - (0.693/5715 yrs)*time
    time = 2230 years

  3. Determining the Activity of a Sample

    Radium-226, which undergoes alpha decay, has a half-life of 1600 years. How many alpha particles are emitted in 1.0 minutes by a 5.0-mg sample of 226Ra, and what is the activity of such a sample in mCi?
    Answer
    1.1 x 1010 alpha particles are emitted
    A = 4.9 mCi
    Explanation
    The sample activity is determined using the expression
    A = kN, where
    A is the activity in units of disintegrations per second
    k is the rate constant in units of s-1
    and N is the number of atoms present
    N can be determined from the mass and molecular weight
    mol = m/MW = (5.0 x 10-3 g)/(226 g/mol) = 2.21 x 10-5 mol
    N = NA * mol = (6.022 x 1023)(2.21 x 10-5 mol)
    N = 1.33 x 1019
    k = 0.693/t1/2 = 0.693/[(1600 yr)*(365 d/yr)*(24 h/d)*(3600 s/h)]
    k = 1.37 x 10-11 s-1
    A = kN = (1.37 x 10-11 s-1)(1.33 x 1019)
    A = 1.827 x 108 disintegrations per second
    Since each disintegration of 226Ra lead to emission of one alpha particle
    A = 1.827 x 108 alphas per second
    In one minute
    A*t = (1.827 x 108 alphas per second)(60 seconds)
    1.1 x 1010 alpha particles are emitted from the sample
    The activity in mCi is determined from the conversion
    1 Ci = 3.7 x 1010 dps
    A = (1.827 x 108 dps)/(3.7 x 1010 dps/Ci)
    A = 4.93 x 10-3 Ci = 4.93 mCi

Topic: Fission and Fusion

  1. Fission and Fusion Reactions

    Complete and balance the nuclear equations for the following fission and fusion reactions.
    • 235U + n -> 160Sm + 72Zn + ...
    • 239Pu + n -> 144Ce + ... + 2n
    • 2H + 2H -> 3H + ...
    • 2H + 2H -> ... + n
    • 233U + n -> 98Nb + ... + 2n
    Answer
    • 4n
    • 94Kr
    • 1H
    • 3He
    • 134Sb
    Explanation
    For the nuclear reaction to be balanced, the number of protons and neutrons must be the same for both products and reactants. Total mass number must be conserved. Therefore,
    235U + n -> 160Sm + 72Zn + ...
    Reactants
    A = 235+1=236, Z = 92+0=92, N =143+1=144

    Products
    A = 160+72=232, Z = 62+30=92, N =98+42=140

    Missing 4 mass units and 4 neutrons from products,
    235U + n -> 160Sm + 72Zn + 4n

    239Pu + n -> 144Ce + ... + 2n
    Reactants
    A = 239+1=240, Z = 94+0=94, N =145+1=146

    Products
    A = 144+2=146, Z = 58+0=58, N =86+2=88

    Missing 94 mass units, 36 nuclear charges, and 58 neutrons from products,
    239Pu + n -> 144Ce + 94Kr + 2n

    2H + 2H -> 3H + ...
    Reactants
    A = 2+2=4, Z = 1+1=2, N = 1+1=2

    Products
    A = 3, Z = 1, N = 2

    Missing 1 mass units, 1 nuclear charges, and 0 neutrons from products,
    2H + 2H -> 3H + 1H

    2H + 2H -> ... + n
    Reactants
    A = 2+2=4, Z = 1+1=2, N = 1+1=2

    Products
    A = 1, Z = 0, N = 1

    Missing 3 mass units, 2 nuclear charges, and 1 neutrons from products,
    2H + 2H -> 3He + n

    233U + n -> 98Nb + ... + 2n
    Reactants
    A = 233+1=234, Z = 92+0=92, N = 141+1=142

    Products
    A = 98+2=100, Z = 41+0=41, N = 57+2=59

    Missing 134 mass units, 51 nuclear charges, and 83 neutrons from products,
    233U + n -> 98Nb + 134Sb + 2n

  2. Energy Release in Fusion Reactions

    Based on the following atomic masses:
    1H = 1.00782 amu

    2H = 2.01410 amu

    3H = 3.01605 amu

    3He = 3.01603 amu

    4He = 4.00260 amu

    and the mass of the neutron (1.00866 amu), calculate the energy released per mole in each of the following nuclear reactions, all of which are possilibities for a controlled fusion process:
    • 2H + 3H -> 4He + n
    • 2H + 2H -> 3He + n
    • 2H + 3He -> 4He + 1H
    Answer
    • 1.70 x 1012 J
    • 3.15 x 1011 J
    • 1.77 x 1012 J
    Explanation
    The energy released in each reaction can be calculated from the mass differences between reactants and products. 2H + 3H -> 4He + n
    DM = M(2H) + M(3H)-[M(4He) + M(n)]
    DM = 2.01410 amu + 3.01605 amu - [4.00260 amu + 1.00866 amu)]
    DM = 0.01889 amu
    For 1 mol of reactants consumed
    DM = 0.01889 g
    The conversion from mass to energy is made using the relation
    E = c2 DM
    E = (2.9979 x 108 m/s)2 (+0.01889 g)(1 kg/ 1000 g)
    1.70 x 1012 J
    2H + 2H -> 3He + n
    DM = M(2H) + M(2H) - [M(3He) + M(n)]
    DM = 2.01410 amu + 2.01410 amu - [3.01603 amu + 1.00866 amu)]
    DM = 0.00351 amu
    For 1 mol of reactants consumed
    DM = 0.00351 g
    The conversion from mass to energy is made using the relation
    E = c2 DM
    E = (2.9979 x 108 m/s)2 (+0.00351 g)(1 kg/ 1000 g)
    3.15 x 1011 J
    2H + 3He -> 4He + 1H
    DM = M(2H) + M(3He)-[M(4He) + M(1H)]
    DM = 2.01410 amu + 3.01603 amu -[4.00260 amu + 1.00782 amu)]
    DM = 0.01971 amu
    For 1 mol of reactants consumed
    DM = 0.01971 g
    The conversion from mass to energy is made using the relation
    E = c2 DM
    E = (2.9979 x 108 m/s)2 (+0.01971 g)(1 kg/ 1000 g)
    1.77 x 1012 J

  3. Consumption of Uranium in a Mega Watt Reactor

    The average energy released in the fission of a single 235U nucleus is about 3 x 10-11 J. If the conversion of this energy to electricity in a nuclear power plant is 40% efficient, what mass of 235U undergoes fission in a year in a plant that produces 1000 MW?
    Answer
    Mass ~ 1 x 103 kg
    Explanation