Nucle

Given:
- $Z = 7$ (protons), $N = A - Z = 14 - 7 = 7$ (neutrons)
- $m\left(\frac{14}{7}\mathrm{N}\right) = 14.00307\,\mathrm{u}$
- $m_p = m_H = 1.007825\,\mathrm{u}$, $m_n = 1.008665\,\mathrm{u}$

Formula:
$$\Delta M = \bigl[Z\,m_H + (A-Z)\,m_n\bigr] - M$$

Step 1: Calculate total mass of constituents.
$$\Sigma m = 7 \times 1.007825 + 7 \times 1.008665$$
$$= 7.054775 + 7.060655 = 14.115430\,\mathrm{u}$$

Step 2: Calculate mass defect.
$$\Delta M = 14.115430 - 14.00307 = 0.11236\,\mathrm{u}$$

Step 3: Convert to energy.
$$\Delta E_b = \Delta M \times 931.5\,\mathrm{MeV/u}$$
$$= 0.11236 \times 931.5 = 104.66\,\mathrm{MeV}$$

Binding energy of $\frac{14}{7}\mathrm{N}$ nucleus $\approx 104.66\,\mathrm{MeV}$.

Given data: $m_H = 1.007825\,\mathrm{u}$, $m_n = 1.008665\,\mathrm{u}$, $1\,\mathrm{u} = 931.5\,\mathrm{MeV/c^2}$

---

For $\frac{56}{26}\mathrm{Fe}$: $Z=26$, $A=56$, $N=30$

Step 1: Total mass of constituents.
$$\Sigma m = 26 \times 1.007825 + 30 \times 1.008665$$
$$= 26.20345 + 30.25995 = 56.46340\,\mathrm{u}$$

Step 2: Mass defect.
$$\Delta M = 56.46340 - 55.934939 = 0.528461\,\mathrm{u}$$

Step 3: Binding energy.
$$\Delta E_b = 0.528461 \times 931.5 = 492.26\,\mathrm{MeV}$$

Binding energy per nucleon $= \dfrac{492.26}{56} \approx 8.79\,\mathrm{MeV/nucleon}$

---

For $\frac{209}{83}\mathrm{Bi}$: $Z=83$, $A=209$, $N=126$

Step 1: Total mass of constituents.
$$\Sigma m = 83 \times 1.007825 + 126 \times 1.008665$$
$$= 83.64948 + 127.09179 = 210.74127\,\mathrm{u}$$

Step 2: Mass defect.
$$\Delta M = 210.74127 - 208.980388 = 1.760882\,\mathrm{u}$$

Step 3: Binding energy.
$$\Delta E_b = 1.760882 \times 931.5 = 1640.26\,\mathrm{MeV}$$

Binding energy per nucleon $= \dfrac{1640.26}{209} \approx 7.85\,\mathrm{MeV/nucleon}$

Given:
- Mass of coin $= 3.0\,\mathrm{g}$
- $A = 63$, $Z = 29$, $N = 34$
- $m\left(\frac{63}{29}\mathrm{Cu}\right) = 62.92960\,\mathrm{u}$
- $m_H = 1.007825\,\mathrm{u}$, $m_n = 1.008665\,\mathrm{u}$
- $N_A = 6.023 \times 10^{23}\,\mathrm{mol}^{-1}$

Step 1: Number of Cu atoms in the coin.
$$n = \frac{3.0}{63} \times 6.023 \times 10^{23} = \frac{3.0 \times 6.023 \times 10^{23}}{63}$$
$$= 2.868 \times 10^{22}\,\text{atoms}$$

Step 2: Mass defect per nucleus.
$$\Sigma m = 29 \times 1.007825 + 34 \times 1.008665$$
$$= 29.22693 + 34.29461 = 63.52154\,\mathrm{u}$$
$$\Delta M = 63.52154 - 62.92960 = 0.59194\,\mathrm{u}$$

Step 3: Binding energy per nucleus.
$$\Delta E_b = 0.59194 \times 931.5 = 551.38\,\mathrm{MeV}$$

Step 4: Total energy for all nuclei in the coin.
$$E_{\text{total}} = 551.38 \times 2.868 \times 10^{22}\,\mathrm{MeV}$$
$$= 1.5814 \times 10^{25}\,\mathrm{MeV}$$

Converting to Joules:
$$E_{\text{total}} = 1.5814 \times 10^{25} \times 1.6 \times 10^{-13}\,\mathrm{J}$$
$$\boxed{E_{\text{total}} \approx 2.53 \times 10^{12}\,\mathrm{J}}$$

Given:
- $A_{\mathrm{Au}} = 197$, $A_{\mathrm{Ag}} = 107$
- Nuclear radius formula: $R = R_0 A^{1/3}$

Step 1: Write the ratio of radii.
$$\frac{R_{\mathrm{Au}}}{R_{\mathrm{Ag}}} = \frac{R_0 (A_{\mathrm{Au}})^{1/3}}{R_0 (A_{\mathrm{Ag}})^{1/3}} = \left(\frac{197}{107}\right)^{1/3}$$

Step 2: Calculate.
$$\frac{197}{107} = 1.8411$$
$$\left(1.8411\right)^{1/3} \approx 1.2285$$

$$\boxed{\frac{R_{\mathrm{Au}}}{R_{\mathrm{Ag}}} \approx 1.23}$$

*(Note: The question states $\frac{107}{79}\mathrm{Au}$ in the OCR but the standard problem uses $\frac{197}{79}\mathrm{Au}$; the calculation above uses $A=197$ for gold as is standard.)*

Given atomic masses:
- $m({}^2_1\mathrm{H}) = 2.014102\,\mathrm{u}$
- $m({}^3_1\mathrm{H}) = 3.016049\,\mathrm{u}$
- $m({}^{12}_6\mathrm{C}) = 12.000000\,\mathrm{u}$
- $m({}^{20}_{10}\mathrm{Ne}) = 19.992439\,\mathrm{u}$
- $m({}^4_2\mathrm{He}) = 4.002603\,\mathrm{u}$
- $m({}^1_1\mathrm{H}) = 1.007825\,\mathrm{u}$

*Note: Since the same number of electrons appear on both sides, atomic masses can be used directly in place of nuclear masses.*

---

(i) ${}^1_1\mathrm{H} + {}^3_1\mathrm{H} \rightarrow {}^2_1\mathrm{H} + {}^2_1\mathrm{H}$

Initial mass:
$$m_i = 1.007825 + 3.016049 = 4.023874\,\mathrm{u}$$

Final mass:
$$m_f = 2 \times 2.014102 = 4.028204\,\mathrm{u}$$

Mass difference:
$$\Delta m = m_i - m_f = 4.023874 - 4.028204 = -0.004330\,\mathrm{u}$$

Q-value:
$$Q = -0.004330 \times 931.5 = -4.034\,\mathrm{MeV}$$

Since Q < 0, the reaction is endothermic.

---

(ii) ${}^{12}_6\mathrm{C} + {}^{12}_6\mathrm{C} \rightarrow {}^{20}_{10}\mathrm{Ne} + {}^4_2\mathrm{He}$

Initial mass:
$$m_i = 2 \times 12.000000 = 24.000000\,\mathrm{u}$$

Final mass:
$$m_f = 19.992439 + 4.002603 = 23.995042\,\mathrm{u}$$

Mass difference:
$$\Delta m = 24.000000 - 23.995042 = 0.004958\,\mathrm{u}$$

Q-value:
$$Q = 0.004958 \times 931.5 = 4.618\,\mathrm{MeV}$$

Since Q > 0, the reaction is exothermic.

Reaction: ${}^{56}_{26}\mathrm{Fe} \rightarrow 2\,{}^{28}_{13}\mathrm{Al}$

Step 1: Calculate Q-value.
$$Q = \bigl[m({}^{56}_{26}\mathrm{Fe}) - 2\,m({}^{28}_{13}\mathrm{Al})\bigr]c^2$$

Step 2: Substitute values.
$$\Delta m = 55.93494 - 2 \times 27.98191$$
$$= 55.93494 - 55.96382 = -0.02888\,\mathrm{u}$$

Step 3: Convert to MeV.
$$Q = -0.02888 \times 931.5 = -26.90\,\mathrm{MeV}$$

Since Q < 0, the fission of ${}^{56}_{26}\mathrm{Fe}$ into two ${}^{28}_{13}\mathrm{Al}$ fragments is not energetically possible (it is endothermic — energy would need to be supplied). This makes physical sense because ${}^{56}_{26}\mathrm{Fe}$ lies near the peak of the binding energy per nucleon curve, so breaking it into lighter fragments does not release energy.

Given:
- Mass of ${}^{239}_{94}\mathrm{Pu} = 1\,\mathrm{kg} = 1000\,\mathrm{g}$
- Molar mass $= 239\,\mathrm{g/mol}$
- Energy per fission $= 180\,\mathrm{MeV}$
- $N_A = 6.023 \times 10^{23}\,\mathrm{mol}^{-1}$

Step 1: Number of atoms in 1 kg.
$$N = \frac{1000}{239} \times 6.023 \times 10^{23}$$
$$= 4.18 \times 10^{21} \times 6.023 \times 10^{23}\,\text{(incorrect grouping — redo)}$$
$$N = \frac{1000 \times 6.023 \times 10^{23}}{239} = \frac{6.023 \times 10^{26}}{239}$$
$$= 2.52 \times 10^{24}\,\text{atoms}$$

Step 2: Total energy released.
$$E = N \times 180\,\mathrm{MeV}$$
$$= 2.52 \times 10^{24} \times 180$$
$$\boxed{E \approx 4.536 \times 10^{26}\,\mathrm{MeV}}$$

In Joules:
$$E = 4.536 \times 10^{26} \times 1.6 \times 10^{-13} = 7.26 \times 10^{13}\,\mathrm{J}$$

Given:
- Mass of deuterium $= 2.0\,\mathrm{kg} = 2000\,\mathrm{g}$
- Molar mass of deuterium $= 2\,\mathrm{g/mol}$
- Energy per fusion event $= 3.27\,\mathrm{MeV}$ (two deuterons fuse)
- Power of lamp $P = 100\,\mathrm{W}$
- $N_A = 6.023 \times 10^{23}\,\mathrm{mol}^{-1}$

Step 1: Number of deuterium nuclei.
$$N = \frac{2000}{2} \times 6.023 \times 10^{23} = 1000 \times 6.023 \times 10^{23} = 6.023 \times 10^{26}$$

Step 2: Number of fusion events.
Each fusion event consumes 2 deuterons:
$$N_{\text{fusions}} = \frac{6.023 \times 10^{26}}{2} = 3.0115 \times 10^{26}$$

Step 3: Total energy released.
$$E = 3.0115 \times 10^{26} \times 3.27\,\mathrm{MeV}$$
$$= 9.848 \times 10^{26}\,\mathrm{MeV}$$
$$= 9.848 \times 10^{26} \times 1.6 \times 10^{-13}\,\mathrm{J}$$
$$= 1.576 \times 10^{14}\,\mathrm{J}$$

Step 4: Time for which lamp glows.
$$t = \frac{E}{P} = \frac{1.576 \times 10^{14}}{100} = 1.576 \times 10^{12}\,\mathrm{s}$$

Converting to years ($1\,\mathrm{year} = 3.154 \times 10^7\,\mathrm{s}$):
$$t = \frac{1.576 \times 10^{12}}{3.154 \times 10^7} \approx 4.99 \times 10^4\,\mathrm{years}$$

$$\boxed{t \approx 5.0 \times 10^4\,\text{years}}$$

Given:
- Radius of each deuteron $r = 2.0\,\mathrm{fm} = 2.0 \times 10^{-15}\,\mathrm{m}$
- Charge of each deuteron $q = e = 1.6 \times 10^{-19}\,\mathrm{C}$
- $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\mathrm{N\,m^2\,C^{-2}}$

Concept: When two deuterons just touch, the distance between their centres is $d = r + r = 2r$. The potential barrier height equals the Coulomb potential energy at this separation.

Step 1: Distance between centres when touching.
$$d = 2r = 2 \times 2.0 \times 10^{-15} = 4.0 \times 10^{-15}\,\mathrm{m}$$

Step 2: Coulomb potential energy.
$$U = \frac{1}{4\pi\varepsilon_0}\cdot\frac{e^2}{d}$$
$$= \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4.0 \times 10^{-15}}$$
$$= \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{4.0 \times 10^{-15}}$$
$$= \frac{2.304 \times 10^{-28}}{4.0 \times 10^{-15}}$$
$$= 5.76 \times 10^{-14}\,\mathrm{J}$$

Step 3: Convert to MeV.
$$U = \frac{5.76 \times 10^{-14}}{1.6 \times 10^{-13}} = 0.36\,\mathrm{MeV}$$

$$\boxed{\text{Height of potential barrier} \approx 360\,\mathrm{keV} = 0.36\,\mathrm{MeV}}$$

Given: $R = R_0 A^{1/3}$, where $R_0 = 1.2\,\mathrm{fm}$.

Step 1: Volume of the nucleus (assumed spherical).
$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$$

Step 2: Mass of the nucleus.
The nucleus contains $A$ nucleons, each of mass approximately $m$ (mass of a proton or neutron $\approx 1.67 \times 10^{-27}\,\mathrm{kg}$):
$$M \approx A \cdot m$$

Step 3: Nuclear density.
$$\rho = \frac{M}{V} = \frac{A \cdot m}{\dfrac{4}{3}\pi R_0^3 A} = \frac{m}{\dfrac{4}{3}\pi R_0^3}$$

Observation: The mass number $A$ cancels completely from the expression. Therefore, $\rho$ is independent of $A$ — it is the same for all nuclei.

Step 4: Numerical estimate.
$$\rho = \frac{1.67 \times 10^{-27}}{\dfrac{4}{3}\pi (1.2 \times 10^{-15})^3}$$
$$= \frac{1.67 \times 10^{-27}}{\dfrac{4}{3}\pi \times 1.728 \times 10^{-45}}$$
$$= \frac{1.67 \times 10^{-27}}{7.238 \times 10^{-45}}$$
$$\approx 2.3 \times 10^{17}\,\mathrm{kg/m^3}$$

This is of the order of $10^{17}\,\mathrm{kg/m^3}$, confirming that nuclear matter density is nearly constant and independent of the mass number $A$.