Nucle — NCERT Solutions

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.