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

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If the radius of a nucleus having 13 protons and 14 neutrons is 3.6 fm, then the ratio of the volume to surface area of nucleus having 53 protons and 72 neutrons is (A) 12 fm (B) 3 fm (C) 6 fm (D) 2 fm

   ›Reveal solutionSolution

   Uses the empirical nuclear-radius formula $R=R_0A^{1/3}$ to find the radius of the second nucleus, then the simple geometric fact that a sphere's volume-to-surface-area ratio is $R/3$.

   Concept and Intuition

   Nuclei are modelled as spheres whose radius grows with mass number as $R = R_0 A^{1/3}$, reflecting the (nearly) constant nuclear density (constant volume per nucleon). Once we know $R_0$ from one nucleus, we can predict the radius of any other nucleus purely from its mass number. The ratio of volume to surface area for any sphere is a purely geometric fact: $\dfrac{V}{S}=\dfrac{\frac{4}{3}\pi R^3}{4\pi R^2}=\dfrac{R}{3}$.

   Step-by-Step Solution

   1. First nucleus: $A_1 = 13+14 = 27$ nucleons, $R_1 = 3.6$ fm. Since $27^{1/3}=3$: $R_0 = R_1/A_1^{1/3} = 3.6/3 = 1.2$ fm. …

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2. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.If the nuclear radius of $^{27}Al$ is 3.6 fermi, the approximate nuclear radius of $^{64}Cu$ in fermi is (A) 2.4 (B) 1.2 (C) 4.8 (D) 3.6

   ›Reveal solutionSolution

   Using $R=R_0A^{1/3}$ and the given Al radius to find $R_0$, the Cu-64 nuclear radius comes out to 4.8 fm.

   Concept and Intuition

   Nuclear radius follows the empirical law $R = R_0 A^{1/3}$, reflecting the fact that nuclear matter has roughly constant density — so volume (and hence $R^3$) scales directly with the number of nucleons $A$. Given one nucleus's radius, we can extract the constant $R_0$ and then predict any other nucleus's radius from its mass number.

   Step-by-Step Solution

   1. For $^{27}Al$: $R_{Al} = R_0(27)^{1/3} = R_0\times3 = 3.6\,fm \Rightarrow R_0 = 1.2\,fm$. …

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3. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.The ratio of the radii of the nuclei of mass numbers 27 and 64 is (A) 3 : 4 (B) 4 : 3 (C) 9 : 16 (D) 16 : 9

   ›Reveal solutionSolution

   Nuclear radius scales as the cube root of mass number; recognizing 27 and 64 as perfect cubes gives a clean 3:4 ratio.

   Concept and Intuition

   Since nuclear density is roughly constant across nuclei, the volume (and hence $R^3$) is proportional to the mass number A, giving $R=R_0A^{1/3}$.

   Step-by-Step Solution

   1. $R_1/R_2=(A_1/A_2)^{1/3}=(27/64)^{1/3}$.
   2. $27=3^3$ and $64=4^3$, so $(27/64)^{1/3}=3/4$. …

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4. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.The density (in $\mathrm{kg\,m^{-3}}$) of nuclear matter is of the order of (A) $10^{21}$ (B) $10^{17}$ (C) $10^{12}$ (D) $10^{8}$

   ›Reveal solutionSolution

   Nuclear density is a constant, extremely large number (independent of which nucleus), of order $10^{17}\,\text{kg m}^{-3}$.

   Concept and Intuition

   Experiments show the nuclear radius follows $R = R_0 A^{1/3}$ with $R_0 \approx 1.2\times10^{-15}\,\text{m}$, where $A$ is the mass number. Since volume $\propto R^3 \propto A$, and mass $\propto A$ (each nucleon has roughly the same mass), the density $\rho = \text{mass}/\text{volume}$ comes out independent of $A$ — every nucleus, light or heavy, has almost the same density. This is a striking fact used to show nuclear matter is incompressible and nucleons are tightly packed.

   Step-by-Step Solution

   1. Take a representative nucleus, e.g., with mass number $A$, radius $R = 1.2\times10^{-15}A^{1/3}\,\text{m}$.
   2. Mass $\approx A \times 1.67\times10^{-27}\,\text{kg}$ (mass of one nucleon).
   3. Volume $= \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi (1.2\times10^{-15})^3 A$.
   4. Compute $(1.2\times10^{-15})^3 \approx 1.73\times10^{-45}\,\text{m}^3$, so volume $\approx \dfrac{4}{3}\pi\times1.73\times10^{-45}A \approx 7.24\times10^{-45}A\,\text{m}^3$. …

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5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If the surface areas of two nucleii are in the ratio $9:49$, then the ratio of their mass numbers is (A) $27:343$ (B) $9:49$ (C) $3:7$ (D) $49:81$

   ›Reveal solutionSolution

   Since nuclear radius goes as $A^{1/3}$, surface area goes as $A^{2/3}$; inverting the given area ratio gives a mass-number ratio of $27:343$.

   Concept and Intuition

   Nuclei are modeled as spheres with radius $R=R_0A^{1/3}$, where $A$ is the mass number — this comes from nuclear density being roughly constant (volume $\propto A$, and volume $\propto R^3$, so $R\propto A^{1/3}$). Surface area of a sphere is $4\pi R^2$, so area $\propto A^{2/3}$.

   Step-by-Step Solution

   1. Write the area ratio in terms of mass numbers: $\dfrac{S_1}{S_2}=\left(\dfrac{A_1}{A_2}\right)^{2/3}=\dfrac{9}{49}$.
   2. Solve for the mass-number ratio by raising both sides to the power $3/2$: $\dfrac{A_1}{A_2}=\left(\dfrac{9}{49}\right)^{3/2}$. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.A nucleus with atomic mass number 'A' produces another nucleus by loosing 2 alpha particles. The volume of the new nucleus is 60 times that of the alpha particle. The atomic mass number A of the original nucleus is (A) 228 (B) 238 (C) 248 (D) 244

   ›Reveal solutionSolution

   Nuclear volume is proportional to mass number (via $R=R_0A^{1/3}$); losing two alpha particles

   removes 8 from A, and matching the volume ratio (60 times an alpha's volume) gives A = 248.

   Concept and Intuition

   Since nuclear radius $R = R_0 A^{1/3}$, nuclear volume $V \propto R^3 \propto A$ — directly

   proportional to the mass number, with the same constant of proportionality for any nucleus

   (including an alpha particle, which has mass number 4).

   Step-by-Step Solution

   1. Emitting 2 alpha particles removes a total mass number of $2\times4=8$ from the original nucleus (mass number A), leaving a daughter nucleus of mass number $A-8$.
   2. Since volume $\propto$ mass number, the ratio of the daughter's volume to an alpha particle's …

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7. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The surface areas of two nucleii are in the ratio 9:25. The mass numbers of the nucleii are in the ratio (A) 27:125 (B) 9:25 (C) 3:5 (D) 1:1

   ›Reveal solutionSolution

   Nuclear radius scales as $R \propto A^{1/3}$, so surface area scales as $A^{2/3}$; invert to find the mass-number ratio. Answer: 27:125.

   Concept and Intuition

   The nuclear radius formula $R = R_0 A^{1/3}$ tells us nuclear volume scales with mass number $A$ (nucleons packed at roughly constant density), and hence radius scales as $A^{1/3}$. Surface area, being proportional to $R^2$, then scales as $A^{2/3}$.

   Step-by-Step Solution

   1. Surface area ratio $= R_1^2:R_2^2 = 9:25$, so $R_1:R_2 = 3:5$.
   2. Since $R \propto A^{1/3}$, $A_1^{1/3}:A_2^{1/3} = 3:5$.
   3. Cubing both sides: $A_1:A_2 = 3^3:5^3 = 27:125$. …

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8. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
   Q.The ratio of the radii of the nuclei $_{29}X^{64}$ and $_{84}Y^{216}$ is (A) $\dfrac{2}{3}$ (B) $\dfrac{8}{15}$ (C) $\dfrac{8}{3}$ (D) $\dfrac{7}{8}$

   ›Reveal solutionSolution

   Nuclear radius scales as $A^{1/3}$; taking the cube root of the mass-number ratio $64/216$ gives $2/3$.

   Concept and Intuition

   Since nucleons are packed at roughly constant density inside a nucleus, the nuclear volume $\propto A$ (mass number), so the radius $R\propto A^{1/3}$. This is why the atomic number (Z) is irrelevant to this ratio — only the mass number matters.

   Step-by-Step Solution

   1. $R = R_0 A^{1/3}$, so $\dfrac{R_X}{R_Y} = \left(\dfrac{A_X}{A_Y}\right)^{1/3} = \left(\dfrac{64}{216}\right)^{1/3}$.
   2. $64 = 4^3$ and $216 = 6^3$. …

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9. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is (A) 64 (B) 81 (C) 100 (D) 125

   ›Reveal solutionSolution

   Nuclear radius scales as $A^{1/3}$; use the given (A, R) pair to fix $R_0$, then solve for the mass number at $R=6$ fm, giving $A=125$.

   Concept and Intuition

   The empirical nuclear radius formula $R = R_0 A^{1/3}$ reflects that nucleons pack at roughly constant density, so nuclear volume (and hence $R^3$) is proportional to the number of nucleons $A$.

   Step-by-Step Solution

   1. $R = R_0 A^{1/3}$. For $A=64$: $64^{1/3}=4$, so $4.8 = R_0\times4 \Rightarrow R_0=1.2$ fm.
   2. For the new nucleus, $6 = 1.2\,A^{1/3} \Rightarrow A^{1/3} = 5$. …

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10. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Choose the correct statement of the following (A) The nuclear density in general, is independent of mass number A. (B) The radius of a nucleus is directly proportional to the mass number A of the nucleus. (C) The binding energy of a nucleus is inversely proportional to its mass defect. (D) Energy is absorbed when heavy nuclei undergo transmutation into light nuclei.

    ›Reveal solutionSolution

    Nuclear density is essentially constant across all nuclei because both mass and volume scale the same way with mass number $A$ — this is the key experimental fact underlying the liquid-drop model.

    Concept and Intuition

    Nuclear radius empirically follows $R = R_0 A^{1/3}$ (nucleons are packed at essentially constant density, like an incompressible fluid drop). Since nuclear mass $m\approx Am_p$ scales as $A$, and volume $V=\tfrac43\pi R^3 \propto (A^{1/3})^3 = A$, the density $\rho = m/V \propto A/A =$ constant, independent of $A$.

    Step-by-Step Solution

    1. Nuclear radius: $R=R_0A^{1/3}$, so volume $V\propto R^3 \propto A$.
    2. Nuclear mass $\propto A$ (roughly $A$ nucleons of similar mass each).
    3. Density $= \dfrac{mass}{volume} \propto \dfrac{A}{A} =$ constant, i.e. independent of $A$ — confirms (A).
    4. (B) is wrong: radius $\propto A^{1/3}$, not $\propto A$.
    5. (C) is wrong: binding energy $BE = (\Delta m)c^2$ is directly proportional to mass defect $\Delta m$, not inversely. …

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11. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.A nucleus has mass number $A_1$ and volume $V_1$. Another nucleus has mass number $A_2$ and volume $V_2$. If the relation between the mass numbers is $A_2=3A_1$, then $\dfrac{V_1}{V_2}=$ (A) $3^{\frac{1}{3}}$ (B) $\left(\dfrac{1}{3}\right)^{\frac{1}{3}}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{\sqrt{3}}$

    ›Reveal solutionSolution

    Nuclear volume is directly proportional to mass number ($V\propto A$ since $R\propto A^{1/3}$), so $V_1/V_2=A_1/A_2=1/3$.

    Concept and Intuition

    A key nuclear physics fact is that nuclear density is (approximately) constant across nuclei — the radius scales as $R\propto A^{1/3}$ so that volume $V\propto R^3\propto A$. This directly implies that a nucleus with 3 times the mass number has 3 times the volume, not 3 times the radius.

    Step-by-Step Solution

    1. Nuclear radius formula: $R = R_0A^{1/3}$.
    2. Nuclear volume: $V = \dfrac43\pi R^3 = \dfrac43\pi R_0^3 A$, i.e. $V \propto A$. …

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12. AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
    Q.Ratio of density of nuclear matter to density of water is at least ____ $(R_0 = 1.2 \times 10^{-15} \, m$ & $m_p = m_n = 1.67 \times 10^{-27} \, kg)$ (A) $2.307 \times 10^{14} \, kg.m^{-3}$ (B) $2.307 \times 10^{17} \, kg.m^{-3}$ (C) $23.07 \times 10^{14} \, kg.m^{-3}$ (D) $23.07 \times 10^{17} \, kg.m^{-3}$

    ›Reveal solutionSolution

    Nuclear density comes out to $\rho_{nucleus}\approx 2.307\times10^{17}\,kg\,m^{-3}$ regardless of mass number; dividing by water's density ($10^3\,kg\,m^{-3}$) gives the ratio $2.307\times10^{14}$.

    Concept and Intuition

    Because the nuclear radius scales as $R = R_0A^{1/3}$, the nuclear volume scales exactly as $A$ (mass number), which cancels the $A$ in the numerator when computing density — nuclear density is essentially the same for every nucleus, a striking and often-tested fact.

    Step-by-Step Solution

    1. Mass of nucleus $\approx A\,m_p$ (taking $m_p\approx m_n$).
    2. Volume $= \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$.
    3. Density $\rho = \dfrac{A\,m_p}{\frac{4}{3}\pi R_0^3 A} = \dfrac{m_p}{\frac{4}{3}\pi R_0^3}$ — the $A$ cancels.
    4. Compute $R_0^3 = (1.2\times10^{-15})^3 = 1.728\times10^{-45}\,m^3$.
    5. $\frac{4}{3}\pi R_0^3 \approx 4.189\times1.728\times10^{-45} \approx 7.238\times10^{-45}\,m^3$. …

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