Although twenty three isotopes of phosphorus are known (all the possible isotopes from 24P up to 46P — Analytical Chemistry Chemistry Question
Radioactive decay
Although twenty three isotopes of phosphorus are known (all the possible isotopes from 24P up to 46P), only 31P, with spin 1/2, is stable and is therefore present at 100% abundance. The half-integer spin and high abundance of 31P make it useful for nuclear magnetic resonance (NMR) spectroscopic studies of bio-molecules, particularly DNA. [1]
Two radioactive isotopes of phosphorus have half-lives that make them useful for scientific experiments. 32P has a half-life of 14.3 days and 33P has a half-life of 25.3 days. Both radioisotopes of phosphorous, 32P and 33P, are beta emitters and they decay as shown in the following nuclear reactions; [2]
^{32}_{15}P → ^{32}_{16}S + beta^-
^{33}_{15}P → ^{33}_{16}S + beta^-
Isotope isotopic masses:
32P: 31.97390727
33P: 32.9717255
32S: 31.97207100
33S: 32.97145876 [3]
Calculate the energy of β particles emitted in the decay reactions of 32P and 33P. [2]
Model Answer
For the decay of 32P the mass difference is:
Δm = 31.97390727 - 31.97207100 = 0.00183627 amu
ΔE = Δm c^2 = (1.0 * 10^-3 kg * 0.00183627 amu / (6.0221 * 10^23 amu)) * (2.9979 * 10^8 m/s)^2 = 2.740456 * 10^-13 J
Therefore, for the decay of 32P:
ΔE = 2.740456 * 10^-13 J * (1 eV / 1.602 * 10^-19 J) = 1.7106 * 10^6 eV. [3, 4]
For the decay of 33P, the mass difference is:
Δm = 32.9717255 - 32.97145876 = 0.00026674 amu
ΔE = (1.0 * 10^-3 kg * 0.00026674 amu / (6.0221 * 10^23 amu)) * (2.9979 * 10^8 m/s)^2 = 3.98084 * 10^-14 J
Therefore, for the decay of 33P:
ΔE = 3.98084 * 10^-14 J * (1 eV / 1.602 * 10^-19 J) = 0.248 * 10^6 eV. [4]
To shield beta radiation usually lead is used. However, secondary emission of X-rays takes place via a process known as Bremsstrahlung in the case of high energy β-emission. Therefore shielding must be accomplished with low density materials, e.g. Plexiglas, Lucite, plastic, wood, or water. During shielding of β-emission from 32P, X-ray photons of λ = 0.1175 nm are produced. Calculate the energy of X-ray photons in eV. [2, 5]
Model Answer
E = h v = h c / λ
E = (6.6261 * 10^-34 J s * 2.9979 * 10^8 m/s) / (0.1175 * 10^-9 m) = 1.691 * 10^-15 J
Energy of X-ray photons in eV:
E = 1.691 * 10^-15 J * (1 eV / 1.602 * 10^-19 J) = 1.055 * 10^4 eV. [4, 6]
Find the mass of 32P which has activity of 0.10 Ci (1 Ci = 3.7·1010 disintegration/s). [5]
Model Answer
A = k N
Activity = 0.10 Ci * (3.7 * 10^10 disintegrations / (s * 1 Ci)) = 3.7 * 10^9 disintegrations / s
Half-life for 32P is 14.3 days = 1.24 * 10^6 s.
k = ln 2 / t_1/2 = ln 2 / (1.24 * 10^6 s) = 5.61 * 10^-7 s^-1
N = A / k = 3.7 * 10^9 s^-1 / (5.61 * 10^-7 s^-1) = 6.6 * 10^15 32P nuclei
n(32P) = 6.6 * 10^15 nuclei / (6.0221 * 10^23 nuclei/mol) = 1.10 * 10^-8 mol
m(32P) = 1.10 * 10^-8 mol * 31.974 g/mol = 3.50 * 10^-7 g. [6, 7]
A sample containing both radioisotopes 32P and 33P has an initial activity of 9136.2 Ci. If the activity decreases to 4569.7 Ci after 14.3 days, calculate the 32P/33P ratio initially present in the sample. [3, 5]
Model Answer
Decay constant for 32P = ln 2 / 14.3 day = 4.85 * 10^-2 day^-1
Decay constant for 33P = ln 2 / 25.3 day = 2.74 * 10^-2 day^-1
Let activity of 32P = A32 and activity of 33P = A33
Initially: A32 + A33 = 9136.2 Ci (Equation I)
After 14.3 days, total activity:
Atotal = A32 * e^(-k32 * t) + A33 * e^(-k33 * t)
Atotal = A32 * e^(-0.0485 * 14.3) + A33 * e^(-0.0274 * 14.3)
Atotal = 0.50 * A32 + 0.675 * A33 = 4569.7 Ci (Equation II)
Solving Equation I and II gives:
A33 = 9.12 Ci and A32 = 9127.3 Ci
Thus, the initial 32P / 33P nuclei ratio is:
N32 / N33 = (A32 * k33) / (A33 * k32) = (9127.3 * 2.74 * 10^-2) / (9.12 * 4.85 * 10^-2) = 5.63 * 10^2. [7, 8]