Q.A gold nucleus is written as $^{197}_{79}\text{Au}$. Find the number of protons and neutrons in this nucleus, and state its mass number and atomic number.

Showing the 12 most recent of 34 on this concept.

1. CBSE 2026Set 55/3/11 markMCQ
   Q.If $r_1$ and $r_2$ are the radii of atomic nuclei of mass numbers $64$ and $27$ respectively, then the value of $\dfrac{r_1}{r_2}$ is : (A) $1$ (B) $\dfrac{4}{3}$ (C) $\dfrac{3}{4}$ (D) $\dfrac{27}{64}$

   ›Reveal solutionSolution

   Nuclear radius scales as the cube root of mass number: $r \propto A^{1/3}$. Taking the ratio gives $\frac{r_1}{r_2} = \left(\frac{64}{27}\right)^{1/3} = \frac{4}{3}$.

   The size of an atomic nucleus is not arbitrary. Experiments—particularly Rutherford scattering and later electron-scattering studies—revealed that nuclear radius follows a beautiful empirical law that connects it directly to how many nucleons (protons and neutrons) are packed inside.

   The key insight is that nuclear matter has roughly constant density. Think of nucleons as incompressible spheres packed together: if you double the number of nucleons, you double the volume, not the radius. Since volume scales as $r^3$, the radius must scale as the cube root of the number of nucleons.

   $r = r_0 A^{1/3}$

   where $A$ is the mass number (total nucleons) and $r_0 \approx 1.2 \text{ fm}$ is a constant.

   Now let's find the ratio of the two nuclear radii.

   1. Write the radius for each nucleus For the nucleus with mass number $A_1 = 64$:

   $r_1 = r_0 (64)^{1/3}$

   For the nucleus with mass number $A_2 = 27$:

   $r_2 = r_0 (27)^{1/3}$

   2. Form the ratio When we divide $r_1$ by $r_2$, the constant $r_0$ cancels:

   $\frac{r_1}{r_2} = \frac{r_0 (64)^{1/3}}{r_0 (27)^{1/3}} = \frac{(64)^{1/3}}{(27)^{1/3}} = \left(\frac{64}{27}\right)^{1/3}$

   3. Recognize perfect cubes

      Both numbers are perfect cubes:

      • $64 = 4^3$
      • $27 = 3^3$

      So:

   $\left(\frac{64}{27}\right)^{1/3} = \left(\frac{4^3}{3^3}\right)^{1/3} = \frac{4}{3}$

   Watch out

   A common mistake is to take the ratio of mass numbers directly, $\frac{64}{27}$, forgetting the cube-root relationship. Nuclear radius is not proportional to mass number itself.

   ✓Final answer

   The correct option is (B) $\dfrac{4}{3}$.

2. CBSE 2026Set V11 markMCQ
   Q.An example for isobars is :

   (a) $^{2}_{1}\text{H}$ and $^{3}_{1}\text{H}$

   (b) $^{2}_{1}\text{H}$ and $^{4}_{2}\text{He}$

   (c) $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$

   (d) $^{3}_{1}\text{H}$ and $^{3}_{2}\text{He}$

   ›Reveal solutionSolution

   Option (d) $^{3}_{1}\text{H}$ and $^{3}_{2}\text{He}$.

   ✓Final answer

   Option (d) $^{3}_{1}\text{H}$ and $^{3}_{2}\text{He}$.

   Isobars are nuclei that have the same mass number $A$ but different atomic number $Z$. Both $^{3}_{1}\text{H}$ (tritium) and $^{3}_{2}\text{He}$ have $A = 3$ while $Z = 1$ and $Z = 2$ respectively, so they are isobars. (Option (a) are isotopes, and (c) is a near-isobar pair but $A = 198$ vs $197$.)

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If mass number of a nucleus is 'A', then its radius will be -

   (i) $R=R_0A^{1/3}$

   (ii) $R=R_0A^{4/3}$

   (iii) $R=R_0A^{3}$

   (iv) $R=R_0A^{2/3}$

   ›Reveal solutionSolution

   Nuclear radius scales as the cube root of the mass number.

   Experimentally the nuclear radius is found to be $R=R_0A^{1/3}$, where $R_0\approx1.2\times10^{-15}$ m is a constant, consistent with nuclear density being roughly the same for all nuclei (volume $\propto A$).

   ✓Final answer

   (i) $R=R_0A^{1/3}$

4. CBSE 2026Set A1 markMCQ
   Q.Which of the following are known as nucleons? (A) Only protons (B) Protons and electrons (C) Protons and neutrons (D) Neutrons and electrons

   ›Reveal solutionSolution

   Nucleons = protons + neutrons, the constituents of the nucleus.

   The atomic nucleus is built from two kinds of particle: positively charged protons and electrically neutral neutrons. Together these are called nucleons. Electrons orbit the nucleus and are not part of it, so any option containing electrons is wrong.

   ✓Final answer

   (C) Protons and neutrons.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.As compared to 12C atom, 14C atom has

   (a) two extra protons and two extra electrons

   (b) two extra protons but no extra electrons

   (c) two extra neutrons and no extra electrons

   (d) two extra neutrons and two extra electrons

   ›Reveal solutionSolution

   Isotopes of the same element have the same number of protons and electrons but different numbers of neutrons.

   Both $^{12}_{6}C$ and $^{14}_{6}C$ have atomic number $Z = 6$, so both have 6 protons and (as neutral atoms) 6 electrons — these are identical between the two.

   Mass number $A =$ number of protons + neutrons.

   For $^{12}C$: neutrons $= 12 - 6 = 6$.

   For $^{14}C$: neutrons $= 14 - 6 = 8$.

   So $^{14}C$ has $8 - 6 = 2$ extra neutrons, and the same (no extra) number of protons and electrons as $^{12}C$.

   ✓Final answer

   (c) two extra neutrons and no extra electrons.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as ____.

   (a) A^2/3

   (b) A^4/3

   (c) A^1/3

   (d) A^5/3

   ›Reveal solutionSolution

   Nuclear radius scales as $R = R_0A^{1/3}$, so the surface area of a (roughly spherical) nucleus, being proportional to $R^2$, scales as $A^{2/3}$.

   Experiments (electron and alpha-particle scattering) show the nuclear radius depends on mass number as

   $R = R_0 A^{1/3}, \quad R_0 \approx 1.2\times10^{-15}\text{ m}$

   This reflects the fact that nuclear density is roughly the same for all nuclei — nuclear volume, $\propto R^3$, is proportional to A (the number of nucleons).

   For a sphere of radius R, the surface area is $S = 4\pi R^2$. Substituting $R = R_0A^{1/3}$:

   $S = 4\pi R_0^2\, A^{2/3} \;\propto\; A^{2/3}$

   ✓Final answer

   (a) The surface area of the nucleus varies as $A^{2/3}$.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following statements is true-

   (a) Proton is a stable particle and neutron is unstable particle.

   (b) Proton is unstable particle and neutron is stable particle.

   (c) Proton and neutron both are stable particles.

   (d) Proton and neutron both are unstable particles.

   ›Reveal solutionSolution

   A free proton is stable (or has an extremely long lifetime); a free neutron is unstable and decays with a mean life of about 15 minutes.

   A free (isolated) proton does not spontaneously decay — its lifetime is, as far as experiments show, effectively infinite (it is stable, or at least stable on cosmological timescales far exceeding the age of the universe). A free neutron, on the other hand, is unstable: it undergoes beta decay,

   $n \rightarrow p + e^- + \bar u_e$

   with a mean lifetime of about 880 s (~15 minutes). Inside most stable nuclei, however, the neutron is stabilised by nuclear binding energy and does not decay.

   ✓Final answer

   (a) Proton is a stable particle and neutron is unstable particle.

8. CBSE 2025Set JS1 mark
   Q.Differentiate between isotopes and isobars.

   ›Reveal solutionSolution

   Isotopes = same $Z$, different $A$ (same element). Isobars = same $A$, different $Z$ (different elements).

   Concept. A nucleus is specified by its atomic number $Z$ (number of protons) and mass number $A$ (protons + neutrons).

   Feature	Isotopes	Isobars
   Atomic number $Z$	same	different
   Mass number $A$	different	same
   Number of neutrons	different	different
   Chemical element	same element	different elements
   Example	$^{1}_{1}H,\ ^{2}_{1}H,\ ^{3}_{1}H$	$^{40}_{18}Ar,\ ^{40}_{20}Ca$

   Because isotopes have the same $Z$, they occupy the same place in the periodic table and have identical chemical properties; isobars are chemically different elements that happen to share the same total nucleon count.

   ✓Final answer

   Isotopes: same atomic number, different mass number (e.g. $^{1}_{1}H,^{2}_{1}H,^{3}_{1}H$). Isobars: same mass number, different atomic number (e.g. $^{40}_{18}Ar,\ ^{40}_{20}Ca$).

9. CBSE 2025Set D1 markMCQ
   Q.What is the main difference between isotopes of the same element? (A) Number of protons (B) Number of neutrons (C) Number of electrons (D) Atomic number

   ›Reveal solutionSolution

   Isotopes differ only in neutron number; their proton number (and hence chemical identity) is the same.

   Isotopes are nuclides of the same element. Being the same element means the same atomic number Z (same number of protons), and in a neutral atom the same number of electrons — so their chemistry is identical.

   What changes is the mass number A, because the number of neutrons N = A − Z differs. For example, hydrogen has isotopes ¹H, ²H (deuterium) and ³H (tritium) with 0, 1 and 2 neutrons respectively.

   ✓Final answer

   (B) Number of neutrons.

10. CBSE 2025Set ANNUAL1 mark
    Q.____________ is credited with the discovery of the neutron.

    ›Reveal solutionSolution

    The neutron was discovered by James Chadwick in 1932, confirming the existence of a neutral nuclear particle predicted earlier by Rutherford.

    James Chadwick discovered the neutron in 1932 by bombarding beryllium with alpha particles and correctly interpreting the resulting neutral, penetrating radiation as a new particle of mass close to the proton's but with no charge — the neutron. This discovery completed the proton-neutron picture of the nucleus.

    ✓Final answer

    James Chadwick is credited with the discovery of the neutron.

11. CBSE 2025Set ANNUAL1 mark
    Q.How is the radius of the nucleus related to the mass number?

    ›Reveal solutionSolution

    Experiments (electron/alpha scattering) show nuclear radius scales with mass number as R = R0 A^(1/3), consistent with nucleons packed at essentially constant density.

    The empirical relation between nuclear radius R and mass number A is:

    R = R0 A^(1/3)

    where R0 is an approximately constant value, about 1.2 x 10^-15 m (1.2 fm), found from scattering experiments.

    This A^(1/3) dependence means the nuclear VOLUME (proportional to R^3, hence to A) grows in direct proportion to the number of nucleons - i.e. all nuclei have very nearly the same nucleon density, which is exactly the reasoning used in question 23 above to show nuclear density is independent of mass number.

    ✓Final answer

    R = R0 A^(1/3), with R0 approx 1.2 fm - the nuclear radius is proportional to the cube root of the mass number.

12. CBSE 2025Set ANNUAL1 mark
    Q.All atoms except ................ contain neutron in their nuclei. (fill in the blank)

    ›Reveal solutionSolution

    The lightest hydrogen isotope, protium, has a nucleus consisting of a single proton and no neutron — every other atom's nucleus contains at least one neutron.

    A nucleus is characterised by mass number $A = Z + N$, where Z is the number of protons and N the number of neutrons. The ordinary hydrogen atom ($^1_1\text{H}$, protium) has $Z = 1$, $A = 1$, so $N = 0$ — its nucleus is a lone proton with no neutron at all. Every other element (starting from helium, $Z=2$) needs at least one neutron in its nucleus to remain bound and stable, since neutrons provide the extra strong-force attraction needed to overcome the mutual Coulomb repulsion of the protons.

    ✓Final answer

    Hydrogen (protium, ¹H).