Chapter 13 Nuclei NCERT Solutions Class 12 Physics

Question 13.1

Obtain the binding energy (in MeV) of a nitrogen nucleus $^{14}_{7}N$, given $m(^{14}_{7}N) = 14.00307 \text{ u}$.

Given Data

Mass of Proton $m_H = 1.007825 \text{ u}$ (includes electron mass to match atomic mass).
Mass of Neutron $m_n = 1.008665 \text{ u}$.
Nucleus: $Z=7$ (Protons), $A=14$ (Mass Number), $N = A-Z = 7$ (Neutrons).

Mass Defect Calculation

Mass defect $\Delta m = [Z m_H + (A-Z) m_n] – m(N)$

$$\Delta m = [7(1.007825) + 7(1.008665)] – 14.00307$$ $$\Delta m = [7.054775 + 7.060655] – 14.00307$$ $$\Delta m = 14.11543 – 14.00307 = 0.11236 \text{ u}$$

Binding Energy

$1 \text{ u} = 931.5 \text{ MeV}$.

$$BE = \Delta m \times 931.5 = 0.11236 \times 931.5 = 104.66334 \text{ MeV}$$

Binding Energy $\approx 104.66 \text{ MeV}$.

Question 13.2

Obtain the binding energy of the nuclei $^{56}_{26}Fe$ and $^{209}_{83}Bi$ in units of MeV from the following data:
$m(^{56}_{26}Fe) = 55.934939 \text{ u}$
$m(^{209}_{83}Bi) = 208.980388 \text{ u}$

Part 1: Iron (Fe-56)

$Z=26, N=30$.

$$\Delta m = [26(1.007825) + 30(1.008665)] – 55.934939$$ $$\Delta m = [26.20345 + 30.25995] – 55.934939$$ $$\Delta m = 56.4634 – 55.934939 = 0.528461 \text{ u}$$ $$BE_{Fe} = 0.528461 \times 931.5 \approx 492.26 \text{ MeV}$$

Part 2: Bismuth (Bi-209)

$Z=83, N=209-83=126$.

$$\Delta m = [83(1.007825) + 126(1.008665)] – 208.980388$$ $$\Delta m = [83.649475 + 127.09179] – 208.980388$$ $$\Delta m = 210.741265 – 208.980388 = 1.760877 \text{ u}$$ $$BE_{Bi} = 1.760877 \times 931.5 \approx 1640.26 \text{ MeV}$$

[Image of Binding Energy per Nucleon Curve]

$BE(Fe) \approx 492.26 \text{ MeV}$
$BE(Bi) \approx 1640.26 \text{ MeV}$

Question 13.3

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $^{63}_{29}Cu$ atoms (of mass 62.92960 u).

1. Binding Energy per Atom

$Z=29, N=34$. $m_{atom} = 62.92960 \text{ u}$.

$$\Delta m = [29(1.007825) + 34(1.008665)] – 62.92960$$ $$\Delta m = [29.226925 + 34.29461] – 62.92960$$ $$\Delta m = 63.521535 – 62.92960 = 0.591935 \text{ u}$$ $$BE_{atom} = 0.591935 \times 931.5 \approx 551.39 \text{ MeV}$$

2. Total Atoms in Coin

$$N = \frac{\text{Mass}}{\text{Molar Mass}} \times N_A = \frac{3.0}{63} \times 6.023 \times 10^{23} \approx 2.868 \times 10^{22}$$

3. Total Energy

$$E_{total} = N \times BE_{atom} = 2.868 \times 10^{22} \times 551.39 \text{ MeV}$$ $$E_{total} \approx 1.58 \times 10^{25} \text{ MeV}$$

Converting to Joules: $1.58 \times 10^{25} \times 1.6 \times 10^{-13} \approx 2.5 \times 10^{12} \text{ J}$.

Total Energy $\approx 1.58 \times 10^{25} \text{ MeV}$.

Question 13.4

Obtain approximately the ratio of the nuclear radii of the gold isotope $^{197}_{79}Au$ and the silver isotope $^{107}_{47}Ag$.

Formula

Nuclear radius $R = R_0 A^{1/3}$.
Ratio $\frac{R_{Au}}{R_{Ag}} = \left( \frac{A_{Au}}{A_{Ag}} \right)^{1/3}$.

Calculation

$$\text{Ratio} = \left( \frac{197}{107} \right)^{1/3} \approx (1.841)^{1/3} \approx 1.23$$

Ratio $\approx 1.23$.

Question 13.5

Determine the Q-value of the following reactions and state whether they are exothermic or endothermic.
(i) $^1_1H + ^3_1H \to ^2_1H + ^2_1H$
(ii) $^{12}_6C + ^{12}_6C \to ^{20}_{10}Ne + ^4_2He$

(i) Reaction 1

Reactants: $^1_1H (1.007825) + ^3_1H (3.016049)$. Mass = 4.023874 u.
Products: $2 \times ^2_1H (2 \times 2.014102)$. Mass = 4.028204 u.
$\Delta m = 4.023874 – 4.028204 = -0.00433 \text{ u}$.
$Q = -0.00433 \times 931.5 = -4.03 \text{ MeV}$.

Since Q is negative, reaction is Endothermic.

(ii) Reaction 2

Reactants: $2 \times ^{12}C (2 \times 12.000000)$. Mass = 24.000000 u.
Products: $^{20}Ne (19.992439) + ^4He (4.002603)$. Mass = 23.995042 u.
$\Delta m = 24.000000 – 23.995042 = 0.004958 \text{ u}$.
$Q = 0.004958 \times 931.5 = 4.62 \text{ MeV}$.

Since Q is positive, reaction is Exothermic.

Question 13.6

Suppose we think of fission of a $^{56}_{26}Fe$ nucleus into two equal fragments, $^{28}_{13}Al$. Is the fission energetically possible? Argue by working out Q of the process.

Calculation

Reaction: $^{56}_{26}Fe \to 2 \times ^{28}_{13}Al$.
$m(Fe) = 55.93494 \text{ u}$.
$m(Al) = 27.98191 \text{ u}$.

$$Q = m(Fe) – 2 \times m(Al)$$ $$Q = 55.93494 – 2(27.98191) = 55.93494 – 55.96382 = -0.02888 \text{ u}$$ $$Q = -0.02888 \times 931.5 \approx -26.9 \text{ MeV}$$

Q is negative. Fission is NOT energetically possible. (Fe-56 is very stable).

Question 13.7

The fission properties of $^{239}_{94}Pu$ are very similar to those of $^{235}_{92}U$. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure $^{239}_{94}Pu$ undergo fission?

1. Number of Atoms

$$N = \frac{1000 \text{ g}}{239 \text{ g/mol}} \times 6.023 \times 10^{23} \approx 2.52 \times 10^{24} \text{ atoms}$$

2. Total Energy

$$E = N \times 180 \text{ MeV} = 2.52 \times 10^{24} \times 180$$ $$E \approx 4.54 \times 10^{26} \text{ MeV}$$

Energy Released $\approx 4.54 \times 10^{26} \text{ MeV}$.

Question 13.8

How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Reaction: $^2_1H + ^2_1H \to ^3_2He + n + 3.27 \text{ MeV}$.

1. Number of Deuterons

$$N = \frac{2000 \text{ g}}{2 \text{ g/mol}} \times 6.023 \times 10^{23} = 6.023 \times 10^{26} \text{ atoms}$$

2. Total Energy

2 atoms release 3.27 MeV. So, $N$ atoms release $\frac{N}{2} \times 3.27$.

$$E = \frac{6.023 \times 10^{26}}{2} \times 3.27 \times 1.6 \times 10^{-13} \text{ J}$$ $$E \approx 1.57 \times 10^{14} \text{ J}$$

3. Time

$$t = \frac{E}{P} = \frac{1.57 \times 10^{14}}{100} = 1.57 \times 10^{12} \text{ s}$$ $$\text{Years} = \frac{1.57 \times 10^{12}}{3.154 \times 10^7} \approx 4.9 \times 10^4 \text{ years}$$

Time $\approx 49,000$ years.

Question 13.9

Calculate the height of the potential barrier for a head on collision of two deuterons. (Assume hard spheres of radius 2.0 fm).

Concept

Potential barrier is the Coulomb repulsion energy when they just touch. Distance between centers $r = 2R = 2 \times 2.0 = 4.0 \text{ fm}$. Charge $q_1 = q_2 = e$.

Calculation

$$V = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4.0 \times 10^{-15}}$$ $$V = \frac{9 \times 2.56 \times 10^{-29}}{4 \times 10^{-15}} = \frac{23.04}{4} \times 10^{-14} \text{ J}$$ $$V = 5.76 \times 10^{-14} \text{ J}$$ $$\text{In keV: } \frac{5.76 \times 10^{-14}}{1.6 \times 10^{-16}} = 360 \text{ keV}$$

Potential Barrier $\approx 360 \text{ keV}$.

Question 13.10

From the relation $R = R_0 A^{1/3}$, show that the nuclear matter density is nearly constant (independent of A).

Proof

Nuclear Mass $M \approx A \times m_n$ (where $m_n$ is nucleon mass).
Volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$.

Density Calculation

$$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A m_n}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m_n}{4\pi R_0^3}$$

Since $m_n$ and $R_0$ are constants, $\rho$ is constant and independent of A.

Result: Nuclear density is constant $\approx 2.3 \times 10^{17} \text{ kg/m}^3$.