Uranium (U, Z = 92) is a naturally occurring radioactive element which occurs as a mixture of 238U ( — Analytical Chemistry Chemistry Question
Uranium
Uranium (U, Z = 92) is a naturally occurring radioactive element which occurs as a mixture of 238U (99.3 %, t1/2 = 4.47×109 y) and 235U (0.7 %, t1/2 = 7.04×108 y). Both radioisotopes are alpha emitters and were created at the time of nucleosynthesis. Their decay is followed by a different sequence of alpha (4He2+) and beta (β–) disintegrations, which lead through successive transmutations of intermediate radioactive products to stable lead isotopes, 206Pb and 207Pb, respectively (Pb, Z = 82). These sequences form two (out of a total of three) so–called radioactive series. Gamma radiation, which appears in various disintegrations, does not affect the transmutations. 235U is less stable than 238U and reacts more easily with thermal neutrons to undergo fission, a fact which makes 235U a suitable fuel for nuclear reactors. The fission reaction is as follows:
235U + n → U* → fission products + 2 – 3 n + 200 MeV / nucleus
Calculate the total number of alpha and beta particles emitted in each of the two complete natural radioactive series (238U → 206Pb and 235U → 207Pb).
Model Answer
alpha decay: X(A,Z) → X(A – 4, Z – 2) + 4He2+ (2p + 2n) (ΔΑ = –4, ΔΖ = –2)
beta decay: X(A,Z) → X(A, Z+1) + β- + νe- (ΔΑ = 0, ΔΖ = +1)
Since changes in the mass number (A) are due to the emission of alpha particles only, in each series we have:
total alpha particles emitted = ΔAtotal / 4
Alpha emission also changes the atomic number (Z) (ΔZ = –2), so the total decrease in Z due to the total alpha particles emitted would be twice their total number. But Z of the final (stable) element of the radioactive series is higher than the expected Z based on alpha emission. This difference in Z is due to the number of beta particles emitted. Thus:
238U → 206Pb: α = ΔA / 4 = (238 – 206) / 4 = 32 / 4 = 8; β = 2α – ΔΖ = 16 – (92 – 82) = 6
235U → 207Pb: α = ΔA / 4 = (235 – 207) / 4 = 28 / 4 = 7; β = 2α – ΔΖ = 14 – (92 – 82) = 4
Explain why in both radioactive series some chemical elements appear more than once.
Model Answer
This occurs when an alpha decay (ΔZ = –2) is followed by two successive beta decays (ΔZ = +2).
Assuming that the initial isotopic abundance (i. e. at the time of nucleosynthesis) was equal for the two uranium isotopes (235U : 238U = 1 : 1), calculate the age of the Earth (i.e., the time that has elapsed since nucleosynthesis).
Model Answer
For each radioisotope of uranium we can write:
N(235U) = N0(235U) exp(– λ235 t) and N(238U) = N0(238U) exp(–λ238 t)
where N is the number of nuclei at time t, N0 at time t = 0 and λ = ln 2 / t1/2 is the disintegration constant.
At t = 0, N0(235U) = N0(238U), then:
exp(–λ238 t) / exp(–λ235 t) = N(238U) / N(235U) = 99.3 / 0.7 = 142
Thus:
λ235 t – λ238 t = ln 142 = 4.95
λ238 = 0.693 / t1/2 = 0.693 / 4.47×10^9 y = 1.55×10^–10 y^–1 (Note: solution uses approx 4.5×10^9 y, yielding λ238 = 1.54×10^–10 y^–1)
λ235 = 0.693 / t1/2 = 0.693 / 7.04×10^8 y = 9.84×10^–10 y^–1 (Note: solution uses approx 7.1×10^8 y, yielding λ235 = 9.76×10^–10 y^–1)
Using the approximated values from the official solution:
t = 4.95 / (9.76 – 1.54)×10^–10 y^–1 = 4.95 / 8.22×10^–10 y^–1 = 6.0×10^9 y
Calculate the amount (in g) of carbon required to release energy equal to the energy released by the complete fission with neutrons of 1g 235U, using to the following oxidation reaction:
C + O2 → CO2 + 393.5 kJ mol–1 (or 4.1eV / molecule)
Model Answer
The energy released by the complete fission of 1g 235U is:
E = (1 / 235) × 6.022×10^23 × 200 MeV = 5.13 × 10^23 MeV
And the energy released upon combustion of 1 g C is:
E = (1 / 12) × 6.022×10^23 × 4.1 eV = 2.06 × 10^23 eV = 2.06 × 10^17 MeV
Hence, the amount of carbon that would release the same amount of energy as the fission of 1 g 235U is:
m = (5.13 × 10^23) / (2.06 × 10^17) = 2.49 × 10^3 kg C (which is equal to 2.49 × 10^6 g C)