Q.(a) Distinguish between isotopes and isobars, giving one example for each.

Showing the 12 most recent of 17 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.The nuclear density is approximately (A) independent of mass number (B) directly proportional to mass number (C) inversely proportional to mass number (D) directly proportional to A^(1/3)

   ›Reveal solutionSolution

   Nuclear radius R = R₀A^(1/3), so volume ∝ A and mass ∝ A, making density independent of A.

   The nuclear radius follows the empirical relation $R = R_0 A^{1/3}$, where $R_0 \approx 1.2$ fm and $A$ is the mass number.

   Volume of the nucleus: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$, so $V \propto A$.

   Mass of the nucleus: $M \approx A \times m_{nucleon}$, so $M \propto A$.

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 mark
   Q.What do you mean by mass defect of a nucleus?

   ›Reveal solutionSolution

   A bound nucleus weighs slightly LESS than the sum of its free, separate nucleons - that missing mass is the mass defect.

   If you add up the masses of Z free protons and (A-Z) free neutrons that would make up a nucleus of mass number A, this sum is always slightly GREATER than the actual measured mass of the bound nucleus. This difference is called the mass defect:

   delta_m = [Z*m_p + (A-Z)*m_n] - M(nucleus)

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 markMCQ
   Q.For mass defect of 0.4% the binding energy of 1 kilogram material is:

   (a) 3.6 × 10^14 ergs

   (b) 3.6 × 10^-14 J

   (c) 3.6 × 10^-14 ergs

   (d) 3.6 × 10^14 J

   ›Reveal solutionSolution

   Using Einstein's mass-energy relation $E = \Delta m\, c^2$ with $\Delta m = 0.4\%$ of 1 kg gives $3.6\times10^{14}\,\text{J}$.

   Mass defect $\Delta m = 0.4\%\ \text{of}\ 1\,\text{kg} = 0.004\,\text{kg}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set JS1 markMCQ
   Q.The energy is emitted when two nuclei of masses $m_1$ and $m_2$ are fused together to make a nucleus of mass $m$. In this process: (A) $(m_1 + m_2) < m$ (B) $(m_1 + m_2) > m$ (C) $(m_1 + m_2) = m$ (D) $m_1 m_2 > m^2$

   ›Reveal solutionSolution

   Energy is released only if some mass disappears; the product mass $m$ is less than $m_1+m_2$, so $(m_1+m_2) > m$ — option (B).

   Concept — mass–energy equivalence. In fusion, two light nuclei combine. If the process releases energy $Q$, that energy comes from a loss of mass (the mass defect $\Delta m$) through Einstein's relation $E = \Delta m\,c^2$.

   Reasoning.

   $\Delta m = (m_1 + m_2) - m > 0 \quad\Rightarrow\quad Q = \Delta m\,c^2 > 0.$ …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The binding energy of a nucleus is equivalent to

   (a) mass of proton

   (b) mass of neutron

   (c) mass of nucleus

   (d) mass defect of nucleus

   ›Reveal solutionSolution

   A nucleus's mass is always slightly less than the sum of the masses of its separate nucleons; this missing mass, the mass defect, is exactly equivalent (via $E=mc^2$) to the binding energy released when the nucleus formed.

   Mass defect: $\Delta m = [Zm_p + (A-Z)m_n] - M_{nucleus}$

   Binding energy is the energy equivalent of this mass defect:

   $E_b = \Delta m\, c^2$

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be

   (a) (3)^(1/3) : 1

   (b) 1:1

   (c) 1:3

   (d) 3:1

   ›Reveal solutionSolution

   Because the nuclear radius scales as R = R0 A^(1/3), the nuclear volume scales exactly as A, so density = mass/volume comes out essentially the same constant for every nucleus, regardless of A.

   Nuclear radius: R = R0 A^(1/3), where R0 approx 1.2 fm is a constant.

   Nuclear volume: V = (4/3) pi R^3 = (4/3) pi R0^3 A

   Since nuclear mass is approximately proportional to A (mass number, roughly A times the nucleon mass), density:

   rho = mass / volume is proportional to A / A = constant

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 mark
   Q.The density of nuclear matter is independent of the size of the nucleus. (T/F)

   ›Reveal solutionSolution

   This statement is True — nuclear density is essentially the same for all nuclei, independent of the mass number (size) of the nucleus.

   The radius of a nucleus with mass number $A$ is given empirically by $R = R_0 A^{1/3}$, where $R_0 \approx 1.2$ fm is a constant. The nuclear volume is then $V \propto R^3 \propto A$, and since the nuclear mass is also proportional to $A$ (each nucleon has roughly the same mass), the density …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set ANNUAL1 markMCQ
   Q.A nucleus ${}_Z X^A$ has mass represented by $M(A,Z)$. If $M_p$ and $M_n$ denote the mass of proton and neutron respectively and $B\cdot E$, the binding energy in MeV, then

   (a) $B\cdot E = M(A,Z) - ZM_p - (A-Z)M_n$

   (b) $B\cdot E = [ZM_p + AM_n - M(A,Z)]C^2$

   (c) $B\cdot E = [Z\cdot M_p + (A-Z)M_n - M(A,Z)]C^2$

   (d) $B\cdot E = [M(A,Z) - ZM_p - (A-Z)M_n]C^2$

   ›Reveal solutionSolution

   Binding energy equals the mass defect (sum of the masses of the free constituent nucleons minus the actual nuclear mass) multiplied by $C^2$; matching this to the options gives option (c).

   Mass defect

   A nucleus $_ZX^A$ contains $Z$ protons and $(A-Z)$ neutrons. If these nucleons existed freely (unbound), their total mass would be $ZM_p + (A-Z)M_n$. The actual measured mass of the bound nucleus, $M(A,Z)$, is less than this because some mass is converted into the energy that binds the nucleus together. This mass difference is the mass defect:

   $\Delta m = ZM_p + (A-Z)M_n - M(A,Z)$

   Binding energy

   By Einstein's mass-energy relation, this "missing" mass corresponds to the binding energy:

   $B\cdot E = \Delta m\, C^2 = \big[ZM_p + (A-Z)M_n - M(A,Z)\big]C^2$

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2025Set ANNUAL1 markMCQ
   Q.If a star converts all the helium (He) nuclei completely into oxygen (O) nuclei, the energy released per oxygen nucleus is (mass of helium nucleus=4.0026a.m.u, mass of oxygen nucleus= 15.9994 a.m.u)

   (a) 7.6 MeV

   (b) 56.12MeV

   (c) 10.24MeV

   (d) 23.9MeV

   ›Reveal solutionSolution

   4 He nuclei fuse into 1 O nucleus; the mass defect times $931.5\,$MeV/u gives the energy released per O nucleus.

   Since $4\times4=16$, the fusion of 4 helium (He-4) nuclei into one oxygen (O-16) nucleus is the reaction implied:

   $4\,^4_2\text{He} \to\ ^{16}_{8}\text{O}$

   Mass of 4 He nuclei $= 4\times4.0026 = 16.0104\,u$. Mass of one O nucleus $=15.9994\,u$.

   $\Delta m = 16.0104 - 15.9994 = 0.0110\,u$ …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set ANNUAL1 mark
    Q.What do you mean by mass defect of a nucleus?

    ›Reveal solutionSolution

    A nucleus always weighs slightly less than the sum of its separate constituent nucleons; this 'missing' mass is the mass defect, and by E = mc^2 it corresponds to the nuclear binding energy.

    For a nucleus $^A_Z X$ made of Z protons and (A-Z) neutrons, the mass defect is defined as

    $\Delta m = \big[Z m_p + (A-Z)m_n\big] - M_{nucleus}$

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set ANNUAL1 mark
    Q.What is the ratio of nuclear densities of two nuclei having mass number $1:4$?

    ›Reveal solutionSolution

    Nuclear density is (almost exactly) the same for all nuclei, regardless of mass number.

    Nuclear density is given by $\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{mA}{\frac43\pi R^3}$, where the nuclear radius scales as $R = R_0 A^{1/3}$. Substituting:

    $\rho = \frac{mA}{\frac43\pi (R_0A^{1/3})^3} = \frac{mA}{\frac43\pi R_0^3 A} = \frac{m}{\frac43\pi R_0^3}$

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2024Set ANNUAL1 markMCQ
    Q.Two nuclei have mass numbers in the ratio 1:3, the ratio of the nuclear densities are

    (a) 3:1

    (b) 1:1

    (c) 1:9

    (d) 1:3

    ›Reveal solutionSolution

    Nuclear density $\rho = \text{mass}/\text{volume} \propto A/A^3 \cdot 3 =$ constant, since the nuclear radius $R\propto A^{1/3}$ makes volume $\propto A$ — so it does not depend on mass number at all.

    The empirical nuclear radius formula is $R = R_0 A^{1/3}$, where $R_0\approx 1.2\,\text{fm}$ is a constant and $A$ is the mass number. The nuclear volume is

    $V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi R_0^3 A$

    which is directly proportional to $A$. The nuclear mass is also (to good approximation) proportional to $A$, since $M \approx A\,u$ (each nucleon contributes roughly one atomic mass unit).

    So the nuclear density is …

    🔒 Start your 14-day free trial to see the full solution →