1. Refer to Moodle . E1. Polonium-210 has a half-life of 138.4 days, decaying by alpha emission. Suppose the helium gas originating from the alpha particles in this decay were collected. What volume of helium at $$ 25^{\circ} \mathrm{C} $$ and 735 mmHg could be obtained from 1.0000 g of polonium dioxide, $$ \mathrm{PoO}_{2} $$, in a period of 48.0 h ? E2. Calculate the binding energy per mole of nucleons for calcium-40, and compare your result with the value in Figure 25.4. Masses needed for this calculation are (in $$ \mathrm{g} / \mathrm{mol} $$ ) $$ { }_{1}^{1} \mathrm{H}=1.00783,{ }_{0}^{1} \mathrm{n}=1.00867 $$, and $$ { }_{20}^{40} \mathrm{Ca}=39.96259 $$. E3. Complete the following nuclear equations. Write the mass number, atomic number, and symbol for the remaining particle. (a) $$ { }_{47}^{111} \mathrm{Ag} \rightarrow{ }^{111} \mathrm{Cd}+ $$ ? (b) $$ { }_{36}^{87} \mathrm{Kr} \rightarrow{ }_{0}^{1} \beta^{+}+? $$ (c) $$ { }_{91}^{231} \mathrm{~Pa} \rightarrow{ }^{227} \mathrm{Ac}+ $$ ? (d) $$ { }_{90}^{230} \mathrm{Th} \rightarrow{ }_{2}^{4} \mathrm{He}+ $$ ? (e) $$ { }_{35}^{82} \mathrm{Br} \rightarrow{ }^{82} \mathrm{Kr}+ $$ ? (f) ? $$ +{ }_{12}^{24} \mathrm{Mg} \rightarrow{ }_{0}^{1} \beta $$ E4. The uranium-235 radioactive decay series, beginning with $$ { }_{92}^{235} \mathrm{U} $$ and ending with $$ { }_{82}^{207} \mathrm{~Pb} $$, occurs in the following sequence: alpha, beta, alpha, beta, alpha, alpha, alpha, alpha, beta, beta, alpha. Write an equation for each step in this series. E5. If a shortage in worldwide supplies of fissionable uranium arose, it would be possible to use other fissionable nuclei. Plutonium, one such fuel, can be made in "breeder" reactors that manufacture more fuel than they consume. The sequence of reactions by which plutonium is made is as follows:

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### E1: Volume of Helium from Polonium-210 Decay

**Known Conditions:**
- Half-life of Polonium-210 ($$ t_{1/2} $$): 138.4 days
- Mass of Polonium dioxide ($$ \text{PoO}_2 $$): 1.0000 g
- Time period: 48.0 h
- Temperature: $$ 25^{\circ} \mathrm{C} $$
- Pressure: 735 mmHg

**Step 1: Calculate the number of moles of $$ \text{PoO}_2 $$**

The molar mass of $$ \text{PoO}_2 $$ can be calculated as follows:
- Molar mass of Polonium (Po): 209 g/mol (approx)
- Molar mass of Oxygen (O): 16 g/mol

$$
\text{Molar mass of } \text{PoO}_2 = 209 + 2 \times 16 = 241 \text{ g/mol}
$$

Now, calculate the number of moles of $$ \text{PoO}_2 $$:

$$
n = \frac{1.0000 \text{ g}}{241 \text{ g/mol}} \approx 0.00415 \text{ mol}
$$

**Step 2: Calculate the decay constant ($$ k $$)**

The decay constant can be calculated using the half-life:

$$
k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{138.4 \text{ days}} \approx 0.00501 \text{ days}^{-1}
$$

Convert $$ k $$ to hours:

$$
k \approx 0.00501 \text{ days}^{-1} \times \frac{1 \text{ day}}{24 \text{ hours}} \approx 0.000209 \text{ h}^{-1}
$$

**Step 3: Calculate the remaining amount of $$ \text{PoO}_2 $$ after 48 hours**

Using the formula for exponential decay:

$$
N(t) = N_0 e^{-kt}
$$

Where:
- $$ N_0 = 0.00415 \text{ mol} $$
- $$ t = 48 \text{ h} $$

$$
N(48) = 0.00415 e^{-0.000209 \times 48} \approx 0.00415 e^{-0.010032} \approx 0.00415 \times 0.9901 \approx 0.00411 \text{ mol}
$$

**Step 4: Calculate the amount of helium produced**

Each decay of $$ \text{Po} $$ produces one helium atom (alpha particle). The number of moles of helium produced is equal to the number of moles of $$ \text{PoO}_2 $$ that decayed:

$$
\Delta N = N_0 - N(48) \approx 0.00415 - 0.00411 = 0.00004 \text{ mol}
$$

**Step 5: Calculate the volume of helium at the given conditions using the ideal gas law**

Using the ideal gas law:

$$
PV = nRT
$$

Where:
- $$ P = 735 \text{ mmHg} $$
- $$ V = ? $$
- $$ n = 0.00004 \text{ mol} $$
- $$ R = 0.0821 \text{ L atm/(K mol)} $$
- $$ T = 25^{\circ}C = 298 \text{ K} $$

Convert pressure to atm:

$$
P = \frac{735 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 0.9671 \text{ atm}
$$

Now, substituting into the ideal gas law:

$$
V = \frac{nRT}{P} = \frac{0.00004 \text{ mol} \times 0.0821 \text{ L atm/(K mol)} \times 298 \text{ K}}{0.9671 \text{ atm}} \approx 0.0001 \text{ L} \approx 0.1 \text{ mL}
$$

### E2: Binding Energy per Mole of Nucleons for Calcium-40

**Known Conditions:**
- Mass of $$ ^{1}_{1}H $$: 1.00783 g/mol
- Mass of $$ ^{0}_{1}n $$: 1.00867 g/mol
- Mass of $$ ^{20}_{40}Ca $$: 39.96259 g/mol

**Step 1: Calculate the mass defect**

The mass defect can be calculated as follows:

$$
\text{Mass defect} = Z \cdot m_H + N \cdot m_n - m_{Ca}
$$

Where:
- $$ Z = 20 $$ (number of protons in $$ ^{40}Ca $$)
- $$ N = 20 $$ (number of neutrons in $$ ^{40}Ca $$)

Calculating the mass defect:

$$
\text{Mass defect} = 20 \cdot 1.00783 + 20 \cdot 1.00867 - 39.96259
$$

Calculating:

$$
= 20.1566 + 20.1734 - 39.96259 \approx 0.36741 \text{ g/mol}
$$

**Step 2: Calculate the binding energy**

The binding energy can be calculated using Einstein's equation $$ E = mc^2 $$:

$$
E = \text{mass defect} \cdot c^2
$$

Where $$ c = 3 \times 10^8 \text{ m/s} $$ and converting mass defect to kg:

$$
\text{mass defect} = 0.36741 \text{ g/mol} = 0.00036741 \text{ kg/mol}
$$

Calculating the binding energy:

$$
E = 0.00036741 \cdot (3 \times 10^8)^2 \approx 0.00036741 \cdot 9 \times 10^{16} \approx 3.30669 \times 10^{13} \text{ J/mol}
$$

**Step 3: Convert to MeV**

1 J = $$ 6.242 \times 10^{12} $$ MeV

$$
E \approx 3.30669 \times 10^{13} \cdot 6.242 \times 10^{12} \approx 206.5 \text{ MeV/mol}
$$

### E3: Completing Nuclear Equations

(a) $$ { }_{47}^{111} \mathrm{Ag} \rightarrow{ }^{111} \mathrm{Cd}+ { }_{0}^{1} \beta^{-} $$

(b) $$ { }_{36}^{87}
(a) $$ { }_{47}^{111} \mathrm{Ag} \rightarrow{ }^{111} \mathrm{Cd}+ { }_{0}^{1} \beta^{-} $$ (b) $$ { }_{36}^{87} \mathrm{Kr} \rightarrow{ }_{0}^{1} \beta^{+}+{ }_{37}^{87} \mathrm{Rb} $$ (c) $$ { }_{91}^{231} \mathrm{Pa} \rightarrow{ }^{227} \mathrm{Ac}+ { }_{4}^{4} \beta^{-} $$ (d) $$ { }_{90}^{230} \mathrm{Th} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{88}^{226} \mathrm{Ra} $$ (e) $$ { }_{35}^{82} \mathrm{Br} \rightarrow{ }^{82} \mathrm{Kr}+ { }_{0}^{1} \beta^{-} $$ (f) $$ { }_{12}^{24} \mathrm{Mg} + { }_{0}^{1} \beta^{+} \rightarrow{ }_{13}^{24} \mathrm{Al} $$

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