Nuclear Physics

Answer: (a)

1. Isobars have the same mass number $A$ but different $Z$.
2. ${}_{1}^{3}\text{H}$ and ${}_{2}^{3}\text{He}$: both $A = 3$, $Z = 1$ and $2$.
3. (c) and (d) are isotopes; (b) has different $A$.
4. So the answer is (a).

Answer: (d)

1. $R = R_0 A^{1/3} \Rightarrow V = \frac{4}{3}\pi R^3 \propto A$: A false, B true.
2. Mass defect $=$ (mass of constituent nucleons) $-$ (mass of nucleus): D true.
3. Atom minus nucleus is just the electrons' mass, not the mass defect: C false.
4. So the answer is (d) B and D are true.

Answer: (d)

1. $R \propto A^{1/3}$, so $\frac{R_\alpha}{R_\beta} = \left(\frac{\alpha}{\beta}\right)^{1/3}$.
2. $= \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} = 0.5$
3. So the answer is (d) 0.5.

Answer: (a)

1. $\rho = \frac{Am}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}$, independent of $A$.
2. So all nuclei have the same density.
3. So the answer is (a) $1 : 1$.

Answer: (c)

1. Electrons do not enter the nucleus; nucleons added $= 10 + 9 = 19$.
2. New $A = 8 + 19 = 27$.
3. Surface area $\propto R^2 \propto A^{2/3}$: $\frac{S_2}{S_1} = \left(\frac{27}{8}\right)^{2/3} = \frac{9}{4}$.
4. Growth $= \frac{9 - 4}{4} \times 100\% = 125\%$
5. So the answer is (c) $125\%$.

Answer: (a)

1. Volume of nucleus $V = \dfrac{M}{\rho} = \dfrac{19.926 \times 10^{-27}}{2.29 \times 10^{17}} = 8.70 \times 10^{-44}\ \text{m}^3$
2. Also $V = \dfrac{4}{3}\pi R_0^3 A = \dfrac{12.56}{3}(1.2\times10^{-15})^3 A = 7.23 \times 10^{-45}A$
3. $A = \dfrac{8.70 \times 10^{-44}}{7.23 \times 10^{-45}} \approx 12$
4. So the answer is (a) 12.

Answer: (b)

1. Mass $\propto A$, volume $\frac{4}{3}\pi R_0^3 A \propto A$.
2. Density $= \frac{\text{mass}}{\text{volume}}$ has no $A$ left.
3. So the answer is (b) independent of A.

Answer: (c)

1. $R = R_0 A^{1/3}$ is the correct relation, so R is correct.
2. It gives $V \propto A$ and mass $\propto A$, so density is the same for Cu and C.
3. So A is false.
4. So the answer is (c) (A) is not correct but (R) is correct.

Answer: (b)

1. $A \propto R^3$, so halving $R$ divides $A$ by $2^3 = 8$.
2. $A = \frac{192}{8} = 24$
3. So the answer is (b) 24.

Answer: 27

1. $A \propto R^3$, so $\frac{A}{64} = \left(\frac{4}{4.8}\right)^3 = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$.
2. $A = \frac{125 \times 64}{216} = \frac{8000}{216} = \frac{1000}{27}$
3. So $x = 27$.

Answer: 4

1. $V = \frac{4}{3}\pi R_0^3 A$, so $V \propto A$.
2. $\frac{V_2}{V_1} = \frac{A_2}{A_1} = 4$
3. So the answer is 4.

Answer: (c)

1. (A) true: Bohr's quantisation $L = n\hbar$.
2. (B) true: nuclear force is short range, not inverse square.
3. (C) true: nuclear force depends on spin.
4. (D) false: nuclear forces are charge independent but non-central.
5. (E) true: smaller packing fraction means a more stable nucleus.
6. So the answer is (c) (A), (B), (C), (E) only.

Answer: (b)

1. Nuclear density $\rho = \frac{3m}{4\pi R_0^3}$ is independent of $A$.
2. So $\rho_O : \rho_{He} = 1 : 1$.
3. So the answer is (b) 1: 1.

Answer: (c)

1. R is true: $R = R_0 A^{1/3}$.
2. From R, $\rho = \frac{3m}{4\pi R_0^3}$ is the same for all nuclei.
3. So the ordering in A is wrong: A is false.
4. So the answer is (c) A is false but R is true.

Answer: 2

1. Momentum conservation: $\dfrac{A_1}{A_2} = \dfrac{v_2}{v_1} = \dfrac{2}{3}$
2. $\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3} = \left(\dfrac{2}{3}\right)^{1/3}$
3. Comparing with $\left(\dfrac{x}{3}\right)^{1/3}$: $x = 2$

Answer: 2

1. $R \propto A^{1/3}$: $\dfrac{R_1}{R_2} = \dfrac{1}{2^{1/3}} \Rightarrow \dfrac{A_1}{A_2} = \dfrac{1}{2}$
2. Momentum conservation: $\dfrac{v_1}{v_2} = \dfrac{A_2}{A_1} = \dfrac{2}{1}$
3. So $n = 2$

Answer: 11

1. $\rho_N = \dfrac{m}{\frac{4}{3}\pi R_0^3} = \dfrac{1.6 \times 10^{-27}}{\frac{4}{3}\pi (1.5 \times 10^{-15})^3}$
2. $\rho_N = \dfrac{1.6 \times 10^{-27}}{1.414 \times 10^{-44}} = 1.13 \times 10^{17}\ \text{kg/m}^3$
3. $\dfrac{\rho_N}{\rho_w} = \dfrac{1.13 \times 10^{17}}{10^3} = 11.3 \times 10^{13}$
4. So $n \approx 11$

Answer: (c)

1. $R \propto A^{1/3}$, so $\frac{R_1}{R_2} = \left(\frac{125}{64}\right)^{1/3}$.
2. $= \frac{5}{4}$
3. So the answer is (c) $5 : 4$.

Answer: (b)

1. $R = R_0 A^{1/3} \Rightarrow \frac{R}{R_0} = A^{1/3}$.
2. Taking log: $\ln\left(\frac{R}{R_0}\right) = \frac{1}{3}\ln A$.
3. This is $y = mx$: a straight line through the origin with slope $\frac{1}{3}$.
4. So the answer is (b), the straight line through O.

Answer: (b)

1. $V \propto R^3 \propto A$: (A) true, (B) false.
2. Mass $\propto A$ and $V \propto A$, so density is independent of $A$: (E) true, (C) and (D) false.
3. So the answer is (b) (A) and (E) only.

Answer: (c)

1. Nuclear density does not depend on mass number.
2. So the ratio is $1 : 1$ whatever the ratio of $A$.
3. So the answer is (c) 1: 1.

Answer: (b)

1. A true: each element has its own characteristic spectrum.
2. B true: Bohr's electron moves in stationary orbits.
3. C false: nuclear density is independent of nuclear size.
4. D false: a free neutron is unstable (half-life about 10 min); a free proton is stable.
5. E true: radioactivity shows nuclear instability.
6. So the answer is (b) A, B and E only.

Answer: (c)

1. Momentum is conserved from rest: $m_1 v_1 = m_2 v_2$.
2. $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{2}$, and $A \propto m$.
3. $R \propto A^{1/3} \Rightarrow \frac{R_1}{R_2} = \left(\frac{1}{2}\right)^{1/3}$
4. So the answer is (c) $\frac{1}{2^{1/3}}$.

Answer: (c)

1. $\rho = \dfrac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \dfrac{m_p}{\frac{4}{3}\pi R_0^3}$, independent of $A$
2. $\rho = \dfrac{1.67 \times 10^{-27}}{\frac{4}{3}\pi (1.3 \times 10^{-15})^3} = \dfrac{1.67 \times 10^{-27}}{9.2 \times 10^{-45}} \approx 1.8 \times 10^{17}\ \text{kg/m}^3$
3. So the answer is (c) $10^{17}\ \text{kg m}^{-3}$.

Answer: (a)

1. Mass $\propto A$ and volume $\propto A$, so nuclear density is the same for all nuclei.
2. $\frac{\rho_{Ca}}{\rho_O} \approx 1$
3. So the answer is (a) 1.

Answer: (c)

1. Parent at rest, so momentum is conserved: $m_1v_1 = m_2v_2 \Rightarrow \dfrac{A_1}{A_2} = \dfrac{v_2}{v_1} = \dfrac{27}{8}$
2. $R \propto A^{1/3} \Rightarrow \dfrac{R_1}{R_2} = \left(\dfrac{27}{8}\right)^{1/3} = \dfrac{3}{2}$
3. So the answer is (c) $3:2$.

Answer: (d)

1. $R \propto A^{1/3}$, so $\frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3}$.
2. $= \frac{5}{3}$
3. So the answer is (d) $\frac{5}{3} R_{Al}$.

Answer: (c)

1. $R \propto A^{1/3}$, so $R_{Cu} = R_{Al}\left(\frac{64}{27}\right)^{1/3}$.
2. $= 3.6 \times \frac{4}{3} = 4.8$ fermi
3. So the answer is (c) 4.8.

Answer: (b)

1. Nuclear density $\rho = \frac{3m}{4\pi R_0^3}$ does not depend on $A$.
2. So the ratio is $1 : 1$ for any mass numbers.
3. So the answer is (b) 1: 1.

Answer: (d)

1. $R \propto A^{1/3}$, so $R_{Te} = 3.6 \times \left(\frac{125}{27}\right)^{1/3}$.
2. $= 3.6 \times \frac{5}{3} = 6.0$ fm
3. So the answer is (d) 6.0 fm.

Answer: (a)

1. $R \propto A^{1/3} \Rightarrow A \propto R^3$.
2. $\frac{A_{Ge}}{9} = 2^3 = 8$
3. $A_{Ge} = 72$
4. So the answer is (a) 72.

Answer: (a)

1. Isotones have the same neutron number $N = A - Z$.
2. Se: $74 - 34 = 40$; Ga: $71 - 31 = 40$, equal.
3. (b) are isotopes, (c) are isobars, (d) have $N = 20$ and $16$.
4. So the answer is (a).

Answer: (b)

1. $R \propto A^{1/3}$, so $R_{Te} = 3.6 \times \left(\frac{125}{27}\right)^{1/3}$.
2. $= 3.6 \times \frac{5}{3} = 6$ fermi
3. So the answer is (b) 6 fermi.

Answer: (b)

1. In ${}_{Z}^{A}X$, $Z$ = number of protons and $A$ = number of nucleons.
2. Neutrons $= A - Z$.
3. So the answer is (b) Z protons and A $-$ Z neutrons.

Answer: (b)

1. Momentum conservation: $m_1v_1 = m_2v_2 \Rightarrow \dfrac{A_1}{A_2} = \dfrac{v_2}{v_1} = \dfrac{1}{2}$
2. $R \propto A^{1/3} \Rightarrow \dfrac{R_1}{R_2} = \left(\dfrac{1}{2}\right)^{1/3}$
3. So the answer is (b) $1 : 2^{1/3}$.

Answer: (c)

1. $A = Z + N$, and $N \ge 0$, so $A \ge Z$; $A$ is never less than $Z$.
2. For ordinary hydrogen ${}_{1}^{1}\text{H}$, $N = 0$, so $A = Z = 1$.
3. So $A$ is sometimes equal to $Z$: the answer is (c).

Answer: (d)

1. Atomic radius $\sim 10^{-10}$ m, nuclear radius $\sim 10^{-15}$ m.
2. Radius ratio $\sim 10^{5}$, and $V \propto r^3$.
3. Volume ratio $\sim (10^{5})^3 = 10^{15}$.
4. So the answer is (d) $10^{15}$.

Answer: (a)

1. Let fraction of ${}^{10}\text{B}$ be $x$: $10x + 11(1 - x) = 10.81$.
2. $11 - x = 10.81 \Rightarrow x = 0.19$
3. Ratio $= 0.19 : 0.81 = 19 : 81$
4. So the answer is (a) $19 : 81$.

Answer: (d)

1. Momentum is conserved from rest: $m_1 v_1 = m_2 v_2$.
2. $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{2}$, and $A \propto m$.
3. $R \propto A^{1/3} \Rightarrow \frac{R_1}{R_2} = \frac{1}{2^{1/3}}$
4. So the answer is (d) $1 : 2^{1/3}$.

Answer: (b)

1. $R \propto A^{1/3}$, so $\frac{R_S}{R_{He}} = \left(\frac{32}{4}\right)^{1/3}$.
2. $= 8^{1/3} = 2$
3. So the answer is (b) 2.

Answer: (c)

1. Mass $\propto A$ and volume $\propto R^3 \propto A$.
2. $\rho = \frac{\text{mass}}{\text{volume}}$ is independent of $A$.
3. So the answer is (c) constant.

Answer: (a)

1. A nucleus is made of nucleons only: protons and neutrons.
2. Electrons lie outside the nucleus.
3. So the answer is (a) neutrons and protons.

Answer: (a)

1. For ${}_{11}\text{Na}^{23}$: $Z = 11$, $A = 23$.
2. Protons $= Z = 11$, neutrons $= A - Z = 12$.
3. Electrons are not in the nucleus, so $0$.
4. So the answer is (a) 11, 12, 0.

Answer: (c)

1. The nuclear force is charge independent.
2. So the nuclear force is the same for p-p, n-n and p-n pairs at the same separation.
3. So the answer is (c) $F_{pp} = F_{nn} = F_{pn}$.

Answer: (a)

1. Neutrons in ${}_{6}\text{C}^{13}$: $13 - 6 = 7$.
2. Neutrons in ${}_{7}\text{N}^{14}$: $14 - 7 = 7$.
3. Same $N$, different $Z$ and $A$, so they are isotones.
4. So the answer is (a) isotones.

Answer: (a)

1. $R \propto A^{1/3}$, so $\frac{R_{Al}}{R_{Te}} = \left(\frac{27}{125}\right)^{1/3}$.
2. $= \frac{3}{5} = \frac{6}{10}$
3. So the answer is (a) $6 : 10$.

Answer: (c)

1. Nuclear forces act only over about 1 to 3 fm and fall to zero beyond that.
2. They follow no simple power law, and they are much stronger than electromagnetic forces.
3. So the answer is (c) they are short range forces.

Answer: (b)

1. $Z = 83$, $N = 209 - 83 = 126$
2. $83 \times 1.007825 + 126 \times 1.008665 = 83.649475 + 127.091790 = 210.741265$ u
3. $\Delta m = 210.741265 - 208.980388 = 1.760877$ u, so $BE = 1.760877 \times 931 = 1639.4$ MeV
4. $\dfrac{BE}{A} = \dfrac{1639.4}{209} = 7.84$ MeV, so the answer is (b).

Answer: (a)

1. Statement I: false. $BE/A$ increases only up to $A \approx 56$ and then decreases for heavier nuclei.
2. Statement II: true. Nuclei move towards greater $BE/A$ (fission of heavy, fusion of light nuclei) because that releases energy and gives more stable nuclei.
3. So the answer is (a).

Answer: (b)

1. ${}_{6}^{12}C$: 6 protons and 6 neutrons.
2. $6(1.00727) + 6(1.00866) = 6.04362 + 6.05196 = 12.09558$ u
3. $\Delta m = 12.09558 - 12 = 0.09558$ u $= 0.09558 \times 931.5 = 89.03$ MeV/$c^2$
4. So the answer is (b) 89.03.

Answer: (a)

1. Stability is judged by binding energy per nucleon. From the reactions: $B_4 = B_3 + 20$ MeV and $B_5 = B_4 - 0.9 = B_3 + 19.1$ MeV.
2. With $B_3 \approx 7.7$ MeV: $\dfrac{B_3}{3} \approx 2.6$, $\dfrac{B_4}{4} \approx 6.9$, $\dfrac{B_5}{5} \approx 5.4$ MeV
3. So $X_4 > X_5 > X_3$, answer (a).

Answer: (a)

1. Assertion: true. For $30 < A < 170$, $BE/A$ stays nearly constant at about 8 MeV.
2. Reason: false. Nuclear force is short range (about 2–3 fm); this short range causes saturation and the constant $BE/A$.
3. So the answer is (a).

Answer: (a)

1. $E = mc^2 = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}$ J
2. $1\ \text{MeV} = 1.6 \times 10^{-13}$ J, so $E = \dfrac{9 \times 10^{13}}{1.6 \times 10^{-13}} = 5.6 \times 10^{26}$ MeV
3. So the answer is (a) $5.6 \times 10^{26}$ MeV.

Answer: (a)

1. Mass defect $\Delta m = \dfrac{BE}{c^2} = \dfrac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8}$ kg
2. $1\ \mu\text{g} = 10^{-9}$ kg, so $\Delta m = 20\ \mu$g
3. So the answer is (a) 20 $\mu$g.

Answer: 1

1. $E = \Delta m\, c^2 = 0.4 \times 10^{-3} \times 9 \times 10^{16} = 3.6 \times 10^{13}$ J
2. $1\ \text{kWh} = 3.6 \times 10^6$ J, so $E = \dfrac{3.6 \times 10^{13}}{3.6 \times 10^6} = 1 \times 10^7$ kWh
3. So $n = 1$

Answer: (d)

1. ${}_{8}\text{O}^{17}$ has $Z = 8$ protons and $N = 17 - 8 = 9$ neutrons.
2. Mass defect $\Delta m = 8M_P + 9M_N - M_O$; binding energy $= \Delta m\, c^2$.
3. Option (d) is $(M_O - 8M_P - 9M_N)c^2$, which has the same magnitude.
4. So the answer is (d).