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

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 down each part of the problem step-by-step:

(a) Binding Energy per Nucleon for Sodium-22

The binding energy per nucleon is the energy required to remove a nucleon from the nucleus. We calculate it by first finding the mass defect and then converting it to energy.

Step 1: Find the mass defect for Na-22.

The mass defect is the difference between the mass of the nucleus and the sum of the masses of the individual nucleons (protons and neutrons).

Mass of Na-22 nucleus = 21.994 437 u
Mass of 11 protons = 11×1.007276470 u=11.07904117 u11 \times 1.007 276 470 \, \text{u} = 11.079 041 17 \, \text{u}11×1.007276470u=11.07904117u
Mass of 11 neutrons = 11×1.008664904 u=11.095314944 u11 \times 1.008 664 904 \, \text{u} = 11.095 314 944 \, \text{u}11×1.008664904u=11.095314944u

Total mass of nucleons = 11.079 041 17 + 11.095 314 944 = 22.174 356 114 u

Mass defect = Total mass of nucleons – Mass of the nucleus
Mass defect = 22.174 356 114 u – 21.994 437 437 u = 0.179 918 677 u

Step 2: Convert the mass defect into energy.

Energy E=Δm⋅c2E = \Delta m \cdot c^2E=Δm⋅c2 (Einstein’s equation), but since we’re working with atomic mass units, we use the conversion factor:

1 atomic mass unit (u) = 931.494 MeV

Energy = 0.179918677 u×931.494 MeV/u=167.78 MeV0.179 918 677 \, \text{u} \times 931.494 \, \text{MeV/u} = 167.78 \, \text{MeV}0.179918677u×931.494MeV/u=167.78MeV

Step 3: Calculate binding energy per nucleon.

Na-22 has 22 nucleons (11 protons + 11 neutrons), so:

Binding energy per nucleon = 167.78 MeV22=7.63 MeV\frac{167.78 \, \text{MeV}}{22} = 7.63 \, \text{MeV}22167.78MeV=7.63MeV

Answer to part (a): The binding energy per nucleon for sodium-22 is 7.63 MeV.

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

Beta plus decay (β+ decay) occurs when a proton is converted into a neutron, releasing a positron and a neutrino. For Na-22:Na1122→Ne1022+β++νe\text{Na}^{22}_{11} \rightarrow \text{Ne}^{22}_{10} + \beta^+ + \nu_eNa1122→Ne1022+β++νe

Where:

Na-22 is sodium-22.
Ne-22 is neon-22.
β+\beta^+β+ is a positron.
νe\nu_eνe is the neutrino.

Answer to part (b): The decay reaction is:Na1122→Ne1022+β++νe\text{Na}^{22}_{11} \rightarrow \text{Ne}^{22}_{10} + \beta^+ + \nu_eNa1122→Ne1022+β++νe

(c) Calculate Q for the Decay Reaction

The Q-value of a decay is the difference in mass between the reactants and products, converted into energy. It can be calculated using:Q=(mNa-22−mNe-22−mβ+−mνe)×c2Q = (m_{\text{Na-22}} – m_{\text{Ne-22}} – m_{\beta^+} – m_{\nu_e}) \times c^2Q=(mNa-22−mNe-22−mβ+−mνe)×c2

Step 1: Find the masses.

Mass of Na-22 = 21.994 437 u
Mass of Ne-22 = 21.991 386 u
Mass of positron (β+) = 0.000 548 580 u
Mass of neutrino (ν) is negligible, so we can ignore it.

Step 2: Calculate the Q-value.

Q=(21.994437 u−21.991386 u−0.000548580 u)×931.494 MeV/uQ = (21.994 437 \, \text{u} – 21.991 386 \, \text{u} – 0.000 548 580 \, \text{u}) \times 931.494 \, \text{MeV/u}Q=(21.994437u−21.991386u−0.000548580u)×931.494MeV/uQ=(0.002502420 u)×931.494 MeV/u=2.33 MeVQ = (0.002 502 420 \, \text{u}) \times 931.494 \, \text{MeV/u} = 2.33 \, \text{MeV}Q=(0.002502420u)×931.494MeV/u=2.33MeV

Answer to part (c): The Q-value for the decay is 2.33 MeV.

(d) Calculate the Mass of Sodium-22 for a Decay Rate of 127,500 Decays per Second

We use the following formula to calculate the mass of Na-22 based on its decay rate:Decay rate=λN\text{Decay rate} = \lambda NDecay rate=λN

Where:

λ\lambdaλ is the decay constant, λ=ln⁡2t12\lambda = \frac{\ln 2}{t_{\frac{1}{2}}}λ=t21ln2
NNN is the number of Na-22 nuclei.

Step 1: Calculate the decay constant.

The half-life of Na-22 is 2.603 years. Convert this to seconds:t12=2.603 years×365.25 days/year×24 hours/day×3600 seconds/hour=8.21×107 secondst_{\frac{1}{2}} = 2.603 \, \text{years} \times 365.25 \, \text{days/year} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} = 8.21 \times 10^7 \, \text{seconds}t21=2.603years×365.25days/year×24hours/day×3600seconds/hour=8.21×107seconds

Now, calculate λ\lambdaλ:λ=ln⁡28.21×107 s=8.44×10−9 s−1\lambda = \frac{\ln 2}{8.21 \times 10^7 \, \text{s}} = 8.44 \times 10^{-9} \, \text{s}^{-1}λ=8.21×107sln2=8.44×10−9s−1

Step 2: Find the number of Na-22 nuclei.

N=Decay rateλ=127,500 decays/s8.44×10−9 s−1=1.51×1013 nucleiN = \frac{\text{Decay rate}}{\lambda} = \frac{127,500 \, \text{decays/s}}{8.44 \times 10^{-9} \, \text{s}^{-1}} = 1.51 \times 10^{13} \, \text{nuclei}N=λDecay rate=8.44×10−9s−1127,500decays/s=1.51×1013nuclei

Step 3: Find the mass.

The number of moles of Na-22 is:moles=NNA=1.51×10136.022×1023=2.51×10−11 moles\text{moles} = \frac{N}{N_A} = \frac{1.51 \times 10^{13}}{6.022 \times 10^{23}} = 2.51 \times 10^{-11} \, \text{moles}moles=NAN=6.022×10231.51×1013=2.51×10−11moles

Now, convert this to mass:mass=moles×molar mass=2.51×10−11 mol×21.994437 g/mol=5.52×10−10 g\text{mass} = \text{moles} \times \text{molar mass} = 2.51 \times 10^{-11} \, \text{mol} \times 21.994 437 \, \text{g/mol} = 5.52 \times 10^{-10} \, \text{g}mass=moles×molar mass=2.51×10−11mol×21.994437g/mol=5.52×10−10g

Answer to part (d): The mass of Na-22 is 5.52 × 10⁻¹⁰ g.

(e) Time for 99.9% of Sodium-22 to Decay

To find the time for 99.9% decay, we use the following formula based on the half-life:N(t)=N0e−λtN(t) = N_0 e^{-\lambda t}N(t)=N0e−λt

Where:

N(t)N(t)N(t) is the amount remaining after time ttt.
N0N_0N0 is the initial amount.
λ\lambdaλ is the decay constant.
ttt is the time.

For 99.9% decay, 0.1% remains, so:0.001N0=N0e−λt0.001 N_0 = N_0 e^{-\lambda t}0.001N0=N0e−λt

Solve for ttt:e−λt=0.001e^{-\lambda t} = 0.001e−λt=0.001−λt=ln⁡(0.001)-\lambda t = \ln(0.001)−λt=ln(0.001)t=ln⁡(0.001)−λ=−6.9078.44×10−9 s−1=8.19×105 st = \frac{\ln(0.001)}{-\lambda} = \frac{-6.907}{8.44 \times 10^{-9} \, \text{s}^{-1}} = 8.19 \times 10^5 \, \text{s}t=−λln(0.001)=8.44×10−9s−1−6.907=8.19×105s

Convert to years:t=8.19×105 s60×60×24×365.25≈25.9 yearst = \frac{8.19 \times 10^5 \, \text{s}}{60 \times 60 \times 24 \times 365.25} \approx 25.9 \, \text{years}t=60×60×24×365.258.19×105s≈25.9years

Answer to part (e): It will take approximately 25.9 years for 99.9% of the sodium-22 to decay.

(a) Binding Energy per Nucleon for Sodium-22

Step 1: Find the mass defect for Na-22.

Step 2: Convert the mass defect into energy.

Step 3: Calculate binding energy per nucleon.

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

(c) Calculate Q for the Decay Reaction

Step 1: Find the masses.

Step 2: Calculate the Q-value.

(d) Calculate the Mass of Sodium-22 for a Decay Rate of 127,500 Decays per Second

Step 1: Calculate the decay constant.

Step 2: Find the number of Na-22 nuclei.

Step 3: Find the mass.

(e) Time for 99.9% of Sodium-22 to Decay

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