NucleiClass 12 Physics NCERT Solutions

To find the binding energy of a nitrogen nucleus (${ }_{7}^{14} \mathrm{N}$), we first calculate its mass defect.

A nitrogen nucleus (${ }_{7}^{14} \mathrm{N}$) contains:

We use the given masses from the textbook data:

Step 1: Calculate the total mass of the constituent nucleons. Total mass of 7 protons and 7 neutrons can be found using the mass of 7 hydrogen atoms and 7 neutrons. This method automatically accounts for the mass of the electrons. Mass of constituents = $(7 \times m_{\mathrm{H}}) + (7 \times m_{\mathrm{n}})$ = $(7 \times 1.007825 \mathrm{u}) + (7 \times 1.008665 \mathrm{u})$ = $7.054775 \mathrm{u} + 7.060655 \mathrm{u}$ = $14.11543 \mathrm{u}$

Step 2: Calculate the mass defect (ΔM). Mass defect is the difference between the total mass of the constituents and the actual atomic mass. ΔM = (Mass of constituents) - (Mass of ${ }_{7}^{14} \mathrm{N}$ atom) ΔM = $14.11543 \mathrm{u} - 14.00307 \mathrm{u}$ ΔM = $0.11236 \mathrm{u}$

Step 3: Calculate the binding energy (E_b). Binding energy is the energy equivalent of the mass defect. We use the conversion factor $1 \mathrm{u} = 931.5 \mathrm{MeV} / c^{2}$. $E_b = \Delta M \times c^2$ $E_b = 0.11236 \times 931.5 \mathrm{MeV}$ $E_b = 104.66334 \mathrm{MeV}$

Thus, the binding energy of the nitrogen nucleus is approximately 104.66 MeV.

We will calculate the binding energy for each nucleus by finding its mass defect and converting it to energy. We use the following data:

Step 1: Calculate the mass of constituents. Mass of constituents = $(26 \times m_{\mathrm{H}}) + (30 \times m_{\mathrm{n}})$ = $(26 \times 1.007825 \mathrm{u}) + (30 \times 1.008665 \mathrm{u})$ = $26.20345 \mathrm{u} + 30.25995 \mathrm{u}$ = $56.46340 \mathrm{u}$

Step 2: Calculate the mass defect (ΔM). ΔM = (Mass of constituents) - (Mass of ${ }_{26}^{56} \mathrm{Fe}$ atom) ΔM = $56.46340 \mathrm{u} - 55.934939 \mathrm{u}$ ΔM = $0.528461 \mathrm{u}$

Step 3: Calculate the binding energy (E_b). $E_b = \Delta M \times 931.5 \mathrm{MeV}$ $E_b = 0.528461 \times 931.5 \mathrm{MeV}$ $E_b = 492.26 \mathrm{MeV}$

Step 1: Calculate the mass of constituents. Mass of constituents = $(83 \times m_{\mathrm{H}}) + (126 \times m_{\mathrm{n}})$ = $(83 \times 1.007825 \mathrm{u}) + (126 \times 1.008665 \mathrm{u})$ = $83.649475 \mathrm{u} + 127.09179 \mathrm{u}$ = $210.741265 \mathrm{u}$

Step 2: Calculate the mass defect (ΔM). ΔM = (Mass of constituents) - (Mass of ${ }_{83}^{209} \mathrm{Bi}$ atom) ΔM = $210.741265 \mathrm{u} - 208.980388 \mathrm{u}$ ΔM = $1.760877 \mathrm{u}$

Step 3: Calculate the binding energy (E_b). $E_b = \Delta M \times 931.5 \mathrm{MeV}$ $E_b = 1.760877 \times 931.5 \mathrm{MeV}$ $E_b = 1640.26 \mathrm{MeV}$

The energy required to separate all the neutrons and protons is equal to the total binding energy of all the nuclei in the coin.

Mass of constituents = $(29 \times m_{\mathrm{H}}) + (34 \times m_{\mathrm{n}})$ = $(29 \times 1.007825 \mathrm{u}) + (34 \times 1.008665 \mathrm{u})$ = $29.226925 \mathrm{u} + 34.29461 \mathrm{u}$ = $63.521535 \mathrm{u}$

Mass defect (ΔM) = (Mass of constituents) - (Mass of ${ }_{29}^{63} \mathrm{Cu}$ atom) ΔM = $63.521535 \mathrm{u} - 62.92960 \mathrm{u}$ ΔM = $0.591935 \mathrm{u}$

Binding energy per nucleus ($E_b$) = $\Delta M \times 931.5 \mathrm{MeV}$ $E_b = 0.591935 \times 931.5 \mathrm{MeV}$ $E_b = 551.39 \mathrm{MeV}$

Number of atoms = (Mass of coin / Molar mass) $\times N_A$ Number of atoms = (3.0 g / 63 g/mol) $\times (6.023 \times 10^{23} \text{ atoms/mol})$ Number of atoms = $2.868 \times 10^{22}$ atoms

Step 3: Calculate the total nuclear energy required. Total Energy = (Binding energy per atom) $\times$ (Number of atoms) Total Energy = $551.39 \mathrm{MeV/atom} \times 2.868 \times 10^{22} \text{ atoms}$ Total Energy = $1.581 \times 10^{25} \mathrm{MeV}$

To express this in Joules, we use the conversion $1 \mathrm{MeV} = 1.6 \times 10^{-13} \mathrm{J}$. Total Energy (in Joules) = $(1.581 \times 10^{25} \mathrm{MeV}) \times (1.6 \times 10^{-13} \mathrm{J/MeV})$ Total Energy (in Joules) = $2.53 \times 10^{12} \mathrm{J}$

Thus, the total nuclear energy required is $2.53 \times 10^{12} \mathrm{J}$.

The radius of a nucleus (R) is related to its mass number (A) by the empirical formula: $R = R_0 A^{1/3}$ where $R_0$ is a constant approximately equal to $1.2 \times 10^{-15} \mathrm{m}$ (1.2 fm).

For the gold isotope (${ }_{79}^{197} \mathrm{Au}$):

For the silver isotope (${ }_{47}^{107} \mathrm{Ag}$):

To find the ratio of their nuclear radii, we divide the radius of the gold nucleus by the radius of the silver nucleus:

The constant $R_0$ cancels out:

Substituting the mass numbers:

$\frac{R_{Au}}{R_{Ag}} = \left( \frac{197}{107} \right)^{1/3}$ $\frac{R_{Au}}{R_{Ag}} = (1.8411)^{1/3}$

Calculating the cube root:

Rounding to two decimal places, the ratio is approximately 1.23.

Thus, the ratio of the nuclear radius of gold to that of silver is approximately 1.23.

The Q-value of a nuclear reaction determines whether energy is released or absorbed. If Q > 0, the reaction is exothermic (releases energy). If Q < 0, the reaction is endothermic (absorbs energy). We will also use the following masses from the textbook data:

Step 1: Calculate the total initial mass. Initial mass = $m\left({ }_{1}^{1} \mathrm{H}\right) + m\left({ }_{1}^{3} \mathrm{H}\right)$ = $1.007825 \mathrm{u} + 3.016049 \mathrm{u}$ = $4.023874 \mathrm{u}$

Step 2: Calculate the total final mass. Final mass = $m\left({ }_{1}^{2} \mathrm{H}\right) + m\left({ }_{1}^{2} \mathrm{H}\right)$ = $2 \times 2.014102 \mathrm{u}$ = $4.028204 \mathrm{u}$

Step 3: Calculate the mass difference (Δm). Δm = Initial mass - Final mass Δm = $4.023874 \mathrm{u} - 4.028204 \mathrm{u}$ Δm = $-0.00433 \mathrm{u}$

Step 4: Calculate the Q-value. $Q = \Delta m \times 931.5 \mathrm{MeV}$ $Q = -0.00433 \times 931.5 \mathrm{MeV}$ $Q = -4.033 \mathrm{MeV}$

Since the Q-value is negative, the reaction is endothermic.

Step 1: Calculate the total initial mass. Initial mass = $m\left({ }_{6}^{12} \mathrm{C}\right) + m\left({ }_{6}^{12} \mathrm{C}\right)$ = $2 \times 12.000000 \mathrm{u}$ = $24.000000 \mathrm{u}$

Step 2: Calculate the total final mass. Final mass = $m\left({ }_{10}^{20} \mathrm{Ne}\right) + m\left({ }_{2}^{4} \mathrm{He}\right)$ = $19.992439 \mathrm{u} + 4.002603 \mathrm{u}$ = $23.995042 \mathrm{u}$

Step 3: Calculate the mass difference (Δm). Δm = Initial mass - Final mass Δm = $24.000000 \mathrm{u} - 23.995042 \mathrm{u}$ Δm = $+0.004958 \mathrm{u}$

Step 4: Calculate the Q-value. $Q = \Delta m \times 931.5 \mathrm{MeV}$ $Q = 0.004958 \times 931.5 \mathrm{MeV}$ $Q = +4.618 \mathrm{MeV}$

Since the Q-value is positive, the reaction is exothermic.

To determine if the fission is energetically possible, we need to calculate the Q-value for the reaction. A reaction is energetically possible if its Q-value is positive, which means energy is released.

The proposed fission reaction is: ${ }_{26}^{56} \mathrm{Fe} \rightarrow { }_{13}^{28} \mathrm{Al} + { }_{13}^{28} \mathrm{Al}$

The Q-value is given by the formula: $Q = (m_{initial} - m_{final}) c^2$

Step 2: Calculate the mass difference (Δm). Δm = $m_{initial} - m_{final}$ Δm = $55.93494 \mathrm{u} - 55.96382 \mathrm{u}$ Δm = $-0.02888 \mathrm{u}$

Step 3: Calculate the Q-value. We convert the mass difference to energy using the conversion factor $1 \mathrm{u} = 931.5 \mathrm{MeV} / c^{2}$. $Q = \Delta m \times 931.5 \mathrm{MeV}$ $Q = -0.02888 \times 931.5 \mathrm{MeV}$ $Q = -26.902 \mathrm{MeV}$

Conclusion: The Q-value for this fission process is -26.902 MeV. Since the Q-value is negative, energy is not released; instead, energy must be supplied to make the reaction happen. Therefore, the fission of a ${ }_{26}^{56} \mathrm{Fe}$ nucleus into two ${ }_{13}^{28} \mathrm{Al}$ fragments is not energetically possible.

This result is consistent with the binding energy curve, which shows that nuclei with mass number around A=56 (like Iron) are among the most stable and have the highest binding energy per nucleon. Fission of such a stable nucleus would lead to less stable products, requiring an input of energy.

To find the total energy released, we first need to determine the number of atoms present in 1 kg of pure ${ }_{94}^{239} \mathrm{Pu}$.

The number of atoms (N) can be calculated using the formula: N = (Total mass / Molar mass) $\times N_A$ N = (1000 g / 239 g/mol) $\times (6.023 \times 10^{23} \text{ atoms/mol})$ N = $4.184 \times (6.023 \times 10^{23})$ atoms N = $2.520 \times 10^{24}$ atoms

Total Energy Released = (Energy per fission) $\times$ (Total number of atoms) Total Energy = $180 \mathrm{MeV/atom} \times (2.520 \times 10^{24} \text{ atoms})$ Total Energy = $453.6 \times 10^{24} \mathrm{MeV}$ Total Energy = $4.536 \times 10^{26} \mathrm{MeV}$

Thus, if all the atoms in 1 kg of pure ${ }_{94}^{239} \mathrm{Pu}$ undergo fission, the total energy released is $4.536 \times 10^{26} \mathrm{MeV}$.

First, we calculate the total energy released from the fusion of 2.0 kg of deuterium.

Number of atoms (N) = (Total mass / Molar mass) $\times N_A$ N = (2000 g / 2 g/mol) $\times (6.023 \times 10^{23} \text{ atoms/mol})$ N = $1000 \times (6.023 \times 10^{23})$ atoms N = $6.023 \times 10^{26}$ atoms

Step 2: Calculate the total energy released. The given reaction shows that 2 deuterium atoms fuse to release 3.27 MeV of energy.

Step 3: Convert the total energy to Joules. We use the conversion factor $1 \mathrm{MeV} = 1.6 \times 10^{-13} \mathrm{J}$. E (in Joules) = $(9.8476 \times 10^{26} \mathrm{MeV}) \times (1.6 \times 10^{-13} \mathrm{J/MeV})$ E = $1.5756 \times 10^{14} \mathrm{J}$

t = $(1.5756 \times 10^{14} \mathrm{J}) / (100 \mathrm{J/s})$ t = $1.5756 \times 10^{12} \mathrm{s}$

Step 5: Convert the time to years (for better perspective). Using the conversion $1 \text{ year} = 3.154 \times 10^{7} \mathrm{s}$. Time (in years) = $(1.5756 \times 10^{12} \mathrm{s}) / (3.154 \times 10^{7} \mathrm{s/year})$ Time = $0.4995 \times 10^{5}$ years Time $\approx 4.995 \times 10^4$ years

Thus, the lamp can be kept glowing for approximately $4.995 \times 10^4$ years.

The height of the potential barrier is the electrostatic potential energy (Coulomb repulsion) when the two deuterons are just touching each other.

The electrostatic potential energy (U) between two charges $q_1$ and $q_2$ separated by a distance r is given by: $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$

Step 2: Calculate the potential energy in Joules. $U = (9 \times 10^9) \times \frac{(1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{4.0 \times 10^{-15}}$ $U = (9 \times 10^9) \times \frac{2.56 \times 10^{-38}}{4.0 \times 10^{-15}}$ $U = (9 \times 0.64) \times 10^{(9 - 38 + 15)}$ $U = 5.76 \times 10^{-14} \mathrm{J}$

Step 3: Convert the energy to electron volts (eV) or kilo-electron volts (keV). The conversion factor is $1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}$. $U(\text{in eV}) = \frac{5.76 \times 10^{-14} \mathrm{J}}{1.6 \times 10^{-19} \mathrm{J/eV}}$ $U(\text{in eV}) = 3.6 \times 10^5 \mathrm{eV}$ $U(\text{in keV}) = 360 \mathrm{keV}$

The height of the potential barrier for a head-on collision of two deuterons is 360 keV.

The density (ρ) of any object is defined as its mass (M) divided by its volume (V). $\rho = \frac{M}{V}$

Step 1: Express the mass of the nucleus. The mass of a nucleus is approximately proportional to its mass number (A). If 'm' is the average mass of a nucleon (proton or neutron), then the mass of the nucleus can be written as: $M \approx m \times A$

Step 2: Express the volume of the nucleus. Assuming the nucleus is spherical, its volume is given by: $V = \frac{4}{3} \pi R^3$ We are given the relation for the nuclear radius: $R = R_0 A^{1/3}$ Substituting this expression for R into the volume formula: $V = \frac{4}{3} \pi (R_0 A^{1/3})^3$ $V = \frac{4}{3} \pi (R_0^3) (A^{(1/3) \times 3})$ $V = \frac{4}{3} \pi R_0^3 A$

Step 3: Calculate the nuclear matter density. Now, we substitute the expressions for mass and volume into the density formula: $\rho = \frac{M}{V} = \frac{m \times A}{\frac{4}{3} \pi R_0^3 A}$

The mass number 'A' appears in both the numerator and the denominator, so it cancels out: $\rho = \frac{m}{\frac{4}{3} \pi R_0^3}$

Conclusion: The resulting expression for density, $\rho = \frac{3m}{4 \pi R_0^3}$, depends only on 'm' (the average mass of a nucleon) and '$R_0$'. Since both 'm' and '$R_0$' are constants, the nuclear matter density (ρ) is also a constant. It does not depend on the mass number A. This shows that the nuclear matter density is nearly constant for all nuclei, regardless of their size.