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The table to the right gives the number of protons \((p)\) and neutrons \((n)\) for four isotopes. (a) Write the symbol for each of the isotopes. (b) Which of the isotopes is most likely to be unstable? (c) Which of the isotopes involves a magic number of protons and/or neutrons? (d) Which isotope will yield potassium-39 following positron emission? \begin{equation}\begin{array}{|c|c|c|c|}\hline & {\text { (i) }} & {\text { (ii) }} & {\text { (iii) }} & {\text { (iv) }} \\ \hline p & {19} & {19} & {20} & {20} \\ \hline n & {19} & {21} & {19} & {20} \\ \hline\end{array} \end{equation}

Short Answer

Expert verified
(a) The symbols for the isotopes are: Isotope (i) - \(K^{38}\), Isotope (ii) - \(K^{40}\), Isotope (iii) - \(Ca^{39}\), and Isotope (iv) - \(Ca^{40}\). (b) Isotope (ii) with 19 protons and 21 neutrons is most likely to be unstable. (c) Isotope (iv) has magic numbers of both protons and neutrons, and Isotope (iii) has a magic number of protons. (d) Isotope (iii) \(Ca^{39}\) will yield potassium-39 following positron emission, as it results in \(K^{39}\).

Step by step solution

01

Writing the symbol for each of the isotopes.

For each isotope, determine the element based on the number of protons and add the mass number as a superscript: - Isotope (i): 19 protons → Potassium (K), K\(^{19+19}\) → \(K^{38}\) - Isotope (ii): 19 protons → Potassium (K), K\(^{19+21}\) → \(K^{40}\) - Isotope (iii): 20 protons → Calcium (Ca), Ca\(^{20+19}\) → \(Ca^{39}\) - Isotope (iv): 20 protons → Calcium (Ca), Ca\(^{20+20}\) → \(Ca^{40}\) Symbols for each isotope: - Isotope (i): \(K^{38}\) - Isotope (ii): \(K^{40}\) - Isotope (iii): \(Ca^{39}\) - Isotope (iv): \(Ca^{40}\)
02

Identify the most likely unstable isotope.

Determine which of these isotopes is likely to be unstable based on unusual proton-neutron ratios. Comparing the isotopes: - Isotope (i): 19 protons and 19 neutrons - Isotope (ii): 19 protons and 21 neutrons - Isotope (iii): 20 protons and 19 neutrons - Isotope (iv): 20 protons and 20 neutrons Isotope (ii) with 19 protons and 21 neutrons has a higher ratio of neutrons to protons compared to other isotopes, so it is the most likely unstable isotope.
03

Identify isotopes with magic numbers.

Magic numbers of protons and neutrons are 2, 8, 20, 28, 50, 82, and 126. Check if any of the isotopes have these magic numbers: - Isotope (i): 19 protons and 19 neutrons (none) - Isotope (ii): 19 protons and 21 neutrons (none) - Isotope (iii): 20 protons (magic) and 19 neutrons (none) - Isotope (iv): 20 protons (magic) and 20 neutrons (magic) Isotope (iv) has magic numbers of both protons and neutrons, and isotope (iii) only has a magic number of protons.
04

Identify the isotope that yields potassium-39 after positron emission.

To find the isotope that yields potassium-39 after positron emission (loses a proton and gains a neutron), look for an isotope that will have 19 protons and a mass number of 39 after this process: - Isotope (i): \(K^{38}\) → no - doesn't have the right mass number - Isotope (ii): \(K^{40}\) → no - doesn't have the right mass number - Isotope (iii): \(Ca^{39}\) → yes, after positron emission, it becomes \(K^{39}\) - Isotope (iv): \(Ca^{40}\) → no - doesn't have the right mass number Therefore, isotope (iii) will yield potassium-39 following positron emission, as it results in \(K^{39}\).

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Most popular questions from this chapter

Write balanced nuclear equations for the following processes: (a) rubidium-90 undergoes beta emission; (b) selenium-72 undergoes electron capture; (c) krypton- 76 undergoes positron emission; (d) radium-226 emits alpha radiation.

Calculations Involving Radioactive Decay and Time If we start with \(1.000 \mathrm{~g}\) of strontium- \(90,0.953 \mathrm{~g}\) will remain after \(2.00 \mathrm{yr}\). (a) What is the half-life of strontium- 90 ? (b) How much strontium- 90 will remain after \(5.00 \mathrm{yr}\) ? (c) What is the initial activity of the sample in becquerels and curies? SOLUTION (a) Analyze We are asked to calculate a half-life, \(t_{1 / 2}\) based on data that tell us how much of a radioactive nucleus has decayed in a time interval \(t=2.00 \mathrm{yr}\) and the information \(N_{0}=1.000 \mathrm{~g}, N_{t}=0.953 \mathrm{~g}\). Plan We first calculate the rate constant for the decay, \(k\), and then use that to compute \(t_{1 / 2}\). Solve Equation \(21.19\) is solved for the decay constant, \(k\), and then Equation \(21.20\) is used to calculate half-life, \(t_{1 / 2}\) : $$ \begin{aligned} k &=-\frac{1}{t} \ln \frac{N_{\mathrm{f}}}{N_{0}}=-\frac{1}{2.00 \mathrm{yr}} \ln \frac{0.953 \mathrm{~g}}{1.000 \mathrm{~g}} \\ &=-\frac{1}{2.00 \mathrm{yr}}(-0.0481)=0.0241 \mathrm{yr}^{-1} \\ t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{0.0241 \mathrm{yr}^{-1}}=28.8 \mathrm{yr} \end{aligned} $$ (b) Analyze We are asked to calculate the amount of a radionuclide remaining after a given period of time. Plan We need to calculate \(N_{b}\) the amount of strontium present at time \(t\), using the initial quantity, \(N_{0}\), and the rate constant for decay, \(k\), calculated in part (a). Solve Again using Equation \(21.19\), with \(k=0.0241 \mathrm{yr}^{-1}\), we have $$ \ln \frac{N_{t}}{N_{0}}=-k t=-\left(0.0241 \mathrm{yr}^{-1}\right)(5.00 \mathrm{yr})=-0.120 $$ \(N_{t} / N_{0}\) is calculated from \(\ln \left(N_{t} / N_{0}\right)=-0.120\) using the \(e^{x}\) or INV LN function of a calculator: $$ \frac{N_{t}}{N_{0}}=e^{-1.120}=0.887 $$ Because \(N_{0}=1.000 \mathrm{~g}\), we have $$ N_{t}=(0.887) N_{0}=(0.887)(1.000 \mathrm{~g})=0.887 \mathrm{~g} $$ (c) Analyze We are asked to calculate the activity of the sample in becquerels and curies. Plan We must calculate the number of disintegrations per atom per second and then multiply by the number of atoms in the sample. Solve The number of disintegrations per atom per second is given by the decay constant, \(k\) : $$ k=\left(\frac{0.0241}{\mathrm{yr}}\right)\left(\frac{1 \mathrm{yr}}{365 \text { days }}\right)\left(\frac{1 \text { day }}{24 \mathrm{~h}}\right)\left(\frac{\mathrm{lh}}{3600 \mathrm{~s}}\right)=7.64 \times 10^{-10} \mathrm{~s}^{-1} $$ To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by \(k\), where we express \(k\) as the number of disintegrations per atom per second, to obtain the number of disintegrations per second: $$ \begin{aligned} \left(1.000 \mathrm{~g}^{90} \mathrm{Sr}\right)\left(\frac{1 \mathrm{~mol}^{90} \mathrm{Sr}}{90 \mathrm{~g}^{90} \mathrm{Sr}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{atoms} \mathrm{Sr}}{1 \mathrm{~mol}^{90} \mathrm{Sr}}\right)=6.7 \times 10^{21} \text { atoms }^{90} \mathrm{Sr} \\ \text { Total disintegrations/s } &=\left(\frac{7.64 \times 10^{-10} \text { disintegrations }}{\text { atom }{ }^{2} \mathrm{~s}}\right)\left(6.7 \times 10^{21} \text { atoms }\right) \\ &=5.1 \times 10^{12} \text { disintegrations/s } \end{aligned} $$ Because \(1 \mathrm{~Bq}\) is one disintegration per second, the activity is \(5.1 \times 10^{12} \mathrm{~Bq}\). The activity in curies is given by $$ \left(5.1 \times 10^{12} \text { disintegrations/s }\right)\left(\frac{1 \mathrm{Ci}}{3.7 \times 10^{10} \text { disintegrations/s }}\right)=1.4 \times 10^{2} \mathrm{Ci} $$ We have used only two significant figures in products of these calculations because we do not know the atomic weight of \({ }^{90} \mathrm{Sr}\) to more than two significant figures without looking it up in a special source.

The nuclear masses of \({ }^{7} \mathrm{Be},{ }^{9} \mathrm{Be}\), and \({ }^{10} \mathrm{Be}\) are \(7.0147,9.0100\), and 10.0113 amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine-131 \(\left(t_{1 / 2}=8.02\right.\) days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a \(1.00-\mu \mathrm{L}\) sample of the water was 214 counts per minutc. After 30 days the level of activity in a \(1.00-\mu \mathrm{I}\). sample was \(15.7\) counts per minute. Did the plants absorb iodide from the water?

How much time is required for a \(6.25-\mathrm{mg}\) sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75 \mathrm{mg}\) if it has a half-life of \(27.8\) days?

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