Sodium-22 (Na-22) is radioactive and decays by beta plus decay mode.

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Sodium-22 (Na-22) is radioactive and decays by beta plus decay mode. The half-life is 2.603 years. (a) Calculate the binding energy per nucleon for sodium-22. (b) Write the reaction for the decay. (c) Calculate Q for the decay reaction. (d) What is the mass of a sample of sodium-22 that has a decay rate of 127,500 decays per second? (e) How long will it take until 99.9% of the sodium has decayed? (Z= 10) Neon-20 19.992 440 u Ne Neon-21 20.993 847 u Neon-22 21.991 386 u (Z = 11) Sodium-22 21.994 437 u Na Sodium-23 22.989 770 u Sodium-24 23.990 963 u (Z= 12) Magnesium-22 21.993 267 u Mg Magnesium-23 22.994 128 u Magnesium-24 23.985 042 u proton 1.007 276 470 u neutron 1.008 664 904 u electron 0.000 548 580 u

The Correct Answer and Explanation is:

Let’s break this down step by step.

(a) Binding Energy per Nucleon for Sodium-22

The binding energy per nucleon is the amount of energy required to remove a nucleon (proton or neutron) from the nucleus, and it can be calculated using the mass defect and Einstein’s equation E=mc2E = mc^2E=mc2.

  1. Mass of Na-22 nucleus: 21.994437 u
  2. Mass of nucleons in Na-22: Sodium-22 consists of 11 protons and 11 neutrons. The total mass of the protons and neutrons is:

Mass of protons=11×1.007276470 u=11.08004117 u\text{Mass of protons} = 11 \times 1.007276470 \, \text{u} = 11.08004117 \, \text{u}Mass of protons=11×1.007276470u=11.08004117uMass of neutrons=11×1.008664904 u=11.09531494 u\text{Mass of neutrons} = 11 \times 1.008664904 \, \text{u} = 11.09531494 \, \text{u}Mass of neutrons=11×1.008664904u=11.09531494uTotal mass of nucleons=11.08004117 u+11.09531494 u=22.17535611 u\text{Total mass of nucleons} = 11.08004117 \, \text{u} + 11.09531494 \, \text{u} = 22.17535611 \, \text{u}Total mass of nucleons=11.08004117u+11.09531494u=22.17535611u

  1. Mass defect:
    The mass defect is the difference between the mass of the nucleus and the total mass of its nucleons:

Mass defect=22.17535611 u−21.994437 u=0.18091911 u\text{Mass defect} = 22.17535611 \, \text{u} – 21.994437 \, \text{u} = 0.18091911 \, \text{u}Mass defect=22.17535611u−21.994437u=0.18091911u

  1. Binding energy:
    Now, we can convert the mass defect into energy using E=mc2E = mc^2E=mc2. Since 1 atomic mass unit (u) corresponds to 931.5 MeV, the binding energy of the nucleus is:

Binding energy=0.18091911 u×931.5 MeV/u=168.23 MeV\text{Binding energy} = 0.18091911 \, \text{u} \times 931.5 \, \text{MeV/u} = 168.23 \, \text{MeV}Binding energy=0.18091911u×931.5MeV/u=168.23MeV

  1. Binding energy per nucleon:
    The number of nucleons in Na-22 is 22 (11 protons + 11 neutrons). Therefore, the binding energy per nucleon is:

Binding energy per nucleon=168.23 MeV22=7.64 MeV\text{Binding energy per nucleon} = \frac{168.23 \, \text{MeV}}{22} = 7.64 \, \text{MeV}Binding energy per nucleon=22168.23MeV​=7.64MeV

(b) Write the Reaction for the Beta-Plus Decay of Sodium-22

Beta-plus decay involves the conversion of a proton into a neutron, emitting a positron and a neutrino. The reaction is:Na-22 (11,22)→Ne-22 (10,22)+e++νe\text{Na-22} \, (11,22) \rightarrow \text{Ne-22} \, (10,22) + e^+ + \nu_eNa-22(11,22)→Ne-22(10,22)+e++νe​

Where:

  • Na-22 is the sodium-22 nucleus (Z=11),
  • Ne-22 is the neon-22 nucleus (Z=10),
  • e+e^+e+ is the emitted positron, and
  • νe\nu_eνe​ is the neutrino.

(c) Calculate Q for the Decay Reaction

The Q-value for a nuclear reaction is the difference in mass between the reactants and products, converted into energy. The general formula is:Q=(Mass of Na-22−Mass of Ne-22−Mass of positron)×c2Q = (\text{Mass of Na-22} – \text{Mass of Ne-22} – \text{Mass of positron}) \times c^2Q=(Mass of Na-22−Mass of Ne-22−Mass of positron)×c2

  1. Mass of Na-22 = 21.994437 u
  2. Mass of Ne-22 = 21.991386 u
  3. Mass of positron = 0.000548580 u

Now, calculate the mass defect:Mass defect=21.994437 u−(21.991386 u+0.000548580 u)=21.994437 u−21.991934580 u=0.00250242 u\text{Mass defect} = 21.994437 \, \text{u} – (21.991386 \, \text{u} + 0.000548580 \, \text{u}) = 21.994437 \, \text{u} – 21.991934580 \, \text{u} = 0.00250242 \, \text{u}Mass defect=21.994437u−(21.991386u+0.000548580u)=21.994437u−21.991934580u=0.00250242u

Convert this mass defect to energy using 1 u=931.5 MeV1 \, \text{u} = 931.5 \, \text{MeV}1u=931.5MeV:Q=0.00250242 u×931.5 MeV/u=2.33 MeVQ = 0.00250242 \, \text{u} \times 931.5 \, \text{MeV/u} = 2.33 \, \text{MeV}Q=0.00250242u×931.5MeV/u=2.33MeV

Thus, the Q-value for the beta-plus decay is 2.33 MeV2.33 \, \text{MeV}2.33MeV.

(d) Decay Rate and Mass of Sodium-22

The decay rate (λ\lambdaλ) is related to the half-life (t1/2t_{1/2}t1/2​) by the formula:λ=ln⁡(2)t1/2\lambda = \frac{\ln(2)}{t_{1/2}}λ=t1/2​ln(2)​

Given the half-life of Na-22 is 2.603 years, we first convert this into seconds:2.603 years=2.603×365.25×24×60×60 seconds=8.213×107 seconds2.603 \, \text{years} = 2.603 \times 365.25 \times 24 \times 60 \times 60 \, \text{seconds} = 8.213 \times 10^7 \, \text{seconds}2.603years=2.603×365.25×24×60×60seconds=8.213×107seconds

Thus:λ=ln⁡(2)8.213×107 s=8.43×10−9 s−1\lambda = \frac{\ln(2)}{8.213 \times 10^7 \, \text{s}} = 8.43 \times 10^{-9} \, \text{s}^{-1}λ=8.213×107sln(2)​=8.43×10−9s−1

Now, the decay rate (RRR) is related to the mass of the sample by:R=λ×NR = \lambda \times NR=λ×N

Where NNN is the number of atoms in the sample, and we can relate this to the mass of the sample mmm using the molar mass of Na-22 (22.994 u). The number of atoms in the sample is:N=mmolar mass×NAN = \frac{m}{\text{molar mass}} \times N_AN=molar massm​×NA​

Where NA=6.022×1023 atoms/molN_A = 6.022 \times 10^{23} \, \text{atoms/mol}NA​=6.022×1023atoms/mol. Solving for mmm:127500=8.43×10−9×m22.994×6.022×1023127500 = 8.43 \times 10^{-9} \times \frac{m}{22.994} \times 6.022 \times 10^{23}127500=8.43×10−9×22.994m​×6.022×1023

Solving for mmm, we get:m=3.12×10−3 kg=3.12 gm = 3.12 \times 10^{-3} \, \text{kg} = 3.12 \, \text{g}m=3.12×10−3kg=3.12g

(e) Time for 99.9% Decay

The formula for the fraction of a substance remaining after time ttt is:N(t)N0=e−λt\frac{N(t)}{N_0} = e^{-\lambda t}N0​N(t)​=e−λt

For 99.9% decay, 0.1% remains, so:0.001=e−λt0.001 = e^{-\lambda t}0.001=e−λt

Taking the natural logarithm of both sides:ln⁡(0.001)=−λt\ln(0.001) = -\lambda tln(0.001)=−λt

Solving for ttt:t=ln⁡(0.001)−λ=−6.907−8.43×10−9=8.2×108 secondst = \frac{\ln(0.001)}{-\lambda} = \frac{-6.907}{-8.43 \times 10^{-9}} = 8.2 \times 10^8 \, \text{seconds}t=−λln(0.001)​=−8.43×10−9−6.907​=8.2×108seconds

Converting this into years:t=8.2×10860×60×24×365.25=26 yearst = \frac{8.2 \times 10^8}{60 \times 60 \times 24 \times 365.25} = 26 \, \text{years}t=60×60×24×365.258.2×108​=26years

Thus, it will take approximately 26 years for 99.9% of the sodium-22 to decay.

Summary of Answers:

  • (a) Binding energy per nucleon: 7.64 MeV7.64 \, \text{MeV}7.64MeV
  • (b) Decay reaction: Na-22→Ne-22+e++νe\text{Na-22} \rightarrow \text{Ne-22} + e^+ + \nu_eNa-22→Ne-22+e++νe​
  • (c) Q-value: 2.33 MeV2.33 \, \text{MeV}2.33MeV
  • (d) Mass of sample: 3.12 g3.12 \, \text{g}3.12g
  • (e) Time for 99.9% decay: 26 years26 \, \text{years}26years

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