Q.Answer the following questions:

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the masses of proton and neutron are $m_p$ and $m_n$ respectively, the experimental masses of $_2He^4$ and $_8O^{16}$ nuclei are $M_1$ and $M_2$ respectively, then (A) $M_2 = 8(m_p + m_n)$ (B) $M_2 < 8(m_p + m_n)$ (C) $M_1 = 2(m_p + m_n)$ (D) $M_1 > 2(m_p + m_n)$

   ›Reveal solutionSolution

   The mass of a stable nucleus is always less than the sum of the masses of its individual protons and neutrons due to the mass defect from binding energy. For both helium-4 and oxygen-16, the experimental mass is less than the sum of the constituent nucleon masses, so the correct choice is (B).

   The key concept here is mass defect and nuclear binding energy. When protons and neutrons bind together to form a nucleus, a small amount of mass is converted into energy (the binding energy) that holds the nucleus together. This means the actual mass of a stable nucleus is always less than the total mass of its separate nucleons. The question tests whether you remember this fundamental principle — not just for one nucleus, but for both.

   Let’s examine each option step by step.

   1. Understand the composition of each nucleus

      • $_2He^4$ has 2 protons and 2 neutrons. If there were no mass defect, its mass would be exactly $2m_p + 2m_n = 2(m_p + m_n)$.
      • $_8O^{16}$ has 8 protons and 8 neutrons. Without mass defect, its mass would be $8m_p + 8m_n = 8(m_p + m_n)$.

   2. Apply the mass defect principle

      For any stable nucleus, the experimental mass is less than the sum of the masses of its free nucleons. This is because the binding energy released during formation corresponds to a loss of mass via $E = \Delta m c^2$.

      • Therefore, for helium-4: $M_1 < 2(m_p + m_n)$.
      • For oxygen-16: $M_2 < 8(m_p + m_n)$.

   3. Evaluate the given options

      • (A) $M_2 = 8(m_p + m_n)$ — This would mean no binding energy, which is false for a real nucleus.
      • (B) $M_2 < 8(m_p + m_n)$ — This matches the mass defect for oxygen-16.
      • (C) $M_1 = 2(m_p + m_n)$ — Again, this would imply no binding energy, false.
      • (D) $M_1 > 2(m_p + m_n)$ — This would mean the nucleus has more mass than its parts, which is impossible for a stable nucleus (it would be energetically unfavorable and would decay). …

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.If E is the kinetic energy of the alpha particle emitted in the decay of a radioactive nucleus of mass number A, then the disintegration energy released in the process is (Parent nucleus is at rest) (A) $\dfrac{(A+4)E}{A}$ (B) $\dfrac{AE}{A-4}$ (C) $\dfrac{(A-4)E}{A}$ (D) $\dfrac{4E}{A-4}$

   ›Reveal solutionSolution

   In alpha decay, the disintegration energy (Q-value) is shared between the alpha particle and the recoiling daughter nucleus as kinetic energy. Using conservation of momentum and energy, the Q-value is found to be $\frac{A E}{A-4}$, which corresponds to option (B).

   The key idea here is that when a parent nucleus at rest emits an alpha particle, the daughter nucleus recoils to conserve momentum. The kinetic energy you measure for the alpha particle is only part of the total energy released — the rest goes into the recoil of the daughter. So the disintegration energy (often called the Q-value) is the sum of both kinetic energies.

   Let’s work through why this leads to the formula in option (B).

   1. Set up the decay process.

      A parent nucleus of mass number $A$ decays by emitting an alpha particle (mass number 4). The daughter nucleus therefore has mass number $A-4$. The parent is initially at rest, so total momentum is zero.

   2. Apply conservation of momentum.

      Let $m_\alpha$ be the mass of the alpha particle and $m_d$ the mass of the daughter. Let $v_\alpha$ and $v_d$ be their speeds after decay. Since initial momentum is zero:

   $m_\alpha v_\alpha = m_d v_d$

   This tells us the daughter recoils in the opposite direction with a speed inversely proportional to its mass.

   3. Relate kinetic energies. The kinetic energy of the alpha particle is $E = \frac{1}{2} m_\alpha v_\alpha^2$. The kinetic energy of the daughter is $E_d = \frac{1}{2} m_d v_d^2$. From momentum conservation, $v_d = \frac{m_\alpha}{m_d} v_\alpha$, so:

   $E_d = \frac{1}{2} m_d \left( \frac{m_\alpha}{m_d} v_\alpha \right)^2 = \frac{1}{2} \frac{m_\alpha^2}{m_d} v_\alpha^2 = \frac{m_\alpha}{m_d} \cdot \frac{1}{2} m_\alpha v_\alpha^2 = \frac{m_\alpha}{m_d} E$

   4. Express masses in terms of mass numbers. For nuclei, mass is approximately proportional to mass number (since nucleon masses are nearly equal). So $m_\alpha \propto 4$ and $m_d \propto A-4$. Thus: …

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3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The range of weak nuclear force is of the order of (A) $10^{16}$ m (B) $10^{-10}$ m (C) $10^{10}$ m (D) $10^{-16}$ m

   ›Reveal solutionSolution

   The weak nuclear force has an extremely short range, roughly $10^{-18}$ m, which is closest to $10^{-16}$ m among the given options. The correct choice is (D).

   The weak nuclear force is one of the four fundamental forces of nature. Unlike gravity or electromagnetism, which have infinite range, the weak force acts only over subatomic distances. This is because it is mediated by very massive particles — the W and Z bosons — whose mass limits how far the force can be felt.

   Why this approach works: The range of a force mediated by a massive particle is inversely proportional to the particle’s mass, given by the uncertainty principle. For the weak force, the mediator mass is about 80–90 GeV/c², which yields a range on the order of $10^{-18}$ m. Comparing this to the options, only $10^{-16}$ m is in the right ballpark (the others are astronomically larger or atomic-scale).

   1. Recall the relation between range and mediator mass

      In quantum field theory, the range $R$ of a force is roughly $R \approx \frac{\hbar}{mc}$, where $m$ is the mass of the force-carrying particle. This comes from the Heisenberg uncertainty principle: a virtual particle of mass $m$ can only exist for a time $\Delta t \sim \hbar/(mc^2)$, and thus travels at most a distance $c \Delta t \sim \hbar/(mc)$.

   2. Plug in the weak boson mass

      The W and Z bosons have masses around $80$–$90$ GeV/c². Using $\hbar c \approx 197$ MeV·fm (where 1 fm = $10^{-15}$ m), we get:

   $R \approx \frac{197 \text{ MeV·fm}}{80\,000 \text{ MeV}} \approx 0.0025 \text{ fm} = 2.5 \times 10^{-18} \text{ m}.$

   3. Compare with the options
      • (A) $10^{16}$ m — larger than the solar system; impossible. …

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4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The range of weak nuclear force is of the order of (A) $10^{16}$ m (B) $10^{-10}$ m (C) $10^{10}$ m (D) $10^{-16}$ m

   ›Reveal solutionSolution

   The weak nuclear force has an extremely short range, roughly $10^{-18}$ m, which is much smaller than the nuclear scale; the closest option given is $10^{-16}$ m, so the answer is (D).

   The weak nuclear force is one of the four fundamental forces of nature. Unlike gravity or electromagnetism, which have infinite range, the weak force acts only over subatomic distances. Why? Because it is mediated by very massive particles — the W and Z bosons — whose mass limits how far they can travel before being absorbed. The range of a force mediated by a massive particle is roughly given by the Compton wavelength of that particle:

   $R \approx \frac{\hbar}{m c}$

   where $m$ is the mass of the mediator. For the W and Z bosons, $m \approx 80\text{–}90 \text{ GeV}/c^2$, which gives a range of about $10^{-18}$ m. That’s about a thousandth the size of a proton’s diameter ($\sim 10^{-15}$ m). So the weak force is indeed very short-ranged.

   Now let’s match this to the options:

   1. Option (A): $10^{16}$ m — This is a light-year scale, far larger than the solar system. No fundamental force except gravity has such a range, and even gravity doesn’t “act” weakly at that scale in the sense of particle physics. Clearly wrong.

   2. Option (B): $10^{-10}$ m — This is the size of an atom (the Bohr radius). The electromagnetic force dominates here; the weak force is negligible at this distance. Wrong.

   3. Option (C): $10^{10}$ m — This is about the distance from Earth to the Sun. Again, not relevant for a subnuclear force. Wrong. …

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5. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.When an element $\ce{^{232}_{90}Th}$ decays into $\ce{^{208}_{82}Pb}$, the number of $\alpha$ and $\beta^-$ particles emitted respectively are (A) 4, 8 (B) 8, 2 (C) 6, 2 (D) 6, 4

   ›Reveal solutionSolution

   The key is to track changes in mass number (A) and atomic number (Z) separately: each α reduces A by 4 and Z by 2; each β⁻ increases Z by 1 without changing A. Solving gives 6 α and 4 β⁻, so the answer is (D).

   Concept & Intuition

   When a heavy nucleus decays to a lighter one, it can emit alpha particles (⁴₂He, which lowers both mass and charge) and beta-minus particles (an electron, which raises the atomic number by converting a neutron to a proton). The mass number only changes via α emission, while the atomic number changes via both α and β⁻. So we can first find the number of α particles from the mass difference, then use the atomic number difference to find the number of β⁻.

   ---

   1. Find the number of α particles from the change in mass number. Thorium-232 has mass number $A = 232$; lead-208 has $A = 208$. Each α particle reduces the mass number by 4. Let $x$ = number of α particles.

   $232 - 4x = 208 \quad \Rightarrow \quad 4x = 24 \quad \Rightarrow \quad x = 6.$

   So 6 α particles are emitted.

   2. Find the net change in atomic number from the α emissions alone. Thorium has $Z = 90$; each α reduces $Z$ by 2. After 6 α emissions, the atomic number would be:

   $90 - 6 \times 2 = 90 - 12 = 78.$

   But lead has $Z = 82$. So we need to increase the atomic number from 78 to 82 — a gain of 4.

   3. Determine the number of β⁻ particles. …

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6. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.In a nuclear fusion reaction, if the mass defect is 0.25%, then the energy released in the fusion of 400 $\mu$g mass of a substance is (A) $9 \times 10^7$ J (B) $9 \times 10^{10}$ J (C) $4.5 \times 10^7$ J (D) $4.5 \times 10^{10}$ J

   ›Reveal solutionSolution

   Only $0.25\%$ of the $400\,\mu$g is converted to energy: $\Delta m=1\,\mu\text{g}=10^{-9}$ kg. Then $E=\Delta m\,c^2=9\times10^{7}$ J (option A).

   Concept. In fusion the mass defect $\Delta m$ is the mass that disappears and reappears as energy, $E=\Delta m\,c^2$. Here the mass defect is $0.25\%$ of the reacting mass.

   Step 1 - Mass converted.

   $\Delta m=\frac{0.25}{100}\times400\,\mu\text{g}=1\,\mu\text{g}.$

   In SI units, $1\,\mu\text{g}=10^{-6}\text{ g}=10^{-9}\text{ kg}$.

   Step 2 - Energy released.

   With $c=3\times10^{8}$ m/s: …

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7. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.If the nucleus $X^{240}$, initially at rest, splits into two daughter nuclei $Y^{100}$ and $Z^{140}$, then the ratio of the kinetic energies of Y and Z is (A) $5:7$ (B) $7:5$ (C) $1:1$ (D) $49:25$

   ›Reveal solutionSolution

   When a nucleus at rest splits into two fragments, their momenta are equal in magnitude and opposite in direction due to conservation of momentum. This leads to their kinetic energies being inversely proportional to their masses, so the ratio of kinetic energies of Y and Z is $\boxed{7:5}$.

   When a heavy nucleus undergoes fission, it splits into two or more lighter nuclei. A fundamental principle governing such processes, especially when the initial system is isolated and at rest, is the conservation of linear momentum. This principle dictates how the resulting fragments will move.

   Here's the core idea: If the parent nucleus is initially at rest, its total momentum is zero. After it splits, the total momentum of the daughter nuclei must still be zero. This means the two daughter nuclei must move in opposite directions with momenta of equal magnitude.

   The relationship between kinetic energy ($K$), momentum ($p$), and mass ($m$) is given by $K = \frac{p^2}{2m}$. Since the magnitudes of the momenta of the two daughter nuclei are equal, their kinetic energies will be inversely proportional to their masses. This inverse relationship is key to solving the problem.

   1. Apply the Principle of Conservation of Linear Momentum:

      The nucleus $X^{240}$ is initially at rest. This means its initial momentum is zero.

      The total initial momentum $P_{initial} = 0$.

      When it splits into two daughter nuclei, $Y^{100}$ and $Z^{140}$, let their momenta be $\vec{p}_Y$ and $\vec{p}_Z$ respectively. According to the conservation of linear momentum, the total momentum after the split must also be zero.

   $\vec{p}_Y + \vec{p}_Z = P_{initial} = 0$

   $\vec{p}_Y = -\vec{p}_Z$

   This equation tells us two things:
   *   The two daughter nuclei move in opposite directions.
   *   The magnitudes of their momenta are equal: $|p_Y| = |p_Z|$. Let's denote this common magnitude as $p$.
   

   2. Relate Kinetic Energy to Momentum and Mass:

   The kinetic energy $K$ of a particle with mass $m$ and momentum $p$ is given by:

   $K = \frac{1}{2}mv^2$

   Since momentum $p = mv$, we can express velocity as $v = p/m$. Substituting this into the kinetic energy formula:
   

   $K = \frac{1}{2}m \left(\frac{p}{m}\right)^2 = \frac{1}{2}m \frac{p^2}{m^2}$

   > [!FORMULA]
   > $$ K = \frac{p^2}{2m} $$
   This form is particularly useful when dealing with problems involving momentum conservation. …
   

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8. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.Which of the following statements regarding nuclear forces is true? (A) The coulomb force between two charges is much stronger than the nuclear force (B) The nuclear force between two nucleons falls rapidly to zero with the increase of the distance between them (C) The nuclear force between neutron-neutron is much stronger than the nuclear force between proton-proton (D) The nuclear force between two charged particles depends on the charges of the particles

   ›Reveal solutionSolution

   Nuclear forces are short-range, charge-independent, and vastly stronger than the Coulomb force at nuclear distances — so only the statement about rapid fall-off with distance is correct.

   The key to this question is understanding what nuclear forces actually are. They are the forces that hold protons and neutrons (collectively called nucleons) together inside the nucleus. Without them, the positively charged protons would fly apart due to electrostatic repulsion. So right away, you can sense that nuclear forces must be much stronger than the Coulomb force at those tiny distances — otherwise, no nucleus could exist.

   Let’s examine each statement one by one.

   1. Statement (A): The Coulomb force between two charges is much stronger than the nuclear force.

      This is false. At the scale of the nucleus (about $10^{-15}$ m), the nuclear force is about 100 times stronger than the electromagnetic repulsion between two protons. That’s why the nucleus stays bound despite the repulsion. The Coulomb force only becomes dominant at larger distances, where the nuclear force has already died out.

   2. Statement (B): The nuclear force between two nucleons falls rapidly to zero with the increase of the distance between them.

      This is true. Nuclear forces have an extremely short range — roughly $1$ to $2$ femtometers ($10^{-15}$ m). Beyond that, the force drops off so steeply that it is effectively zero. This is why nuclei have a finite size and why nucleons only interact with their nearest neighbours inside the nucleus.

   3. Statement (C): The nuclear force between neutron-neutron is much stronger than the nuclear force between proton-proton. …

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9. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.Nuclear fission and fusion can be explained on the basis of (A) Einstein’s theory of relativity (B) Einstein specific heat equation (C) Einstein mass-energy equation (D) Einstein photoelectric equation

   ›Reveal solutionSolution

   The key idea is that both nuclear fission and fusion involve a change in mass, which is converted into energy according to Einstein’s mass-energy equivalence. The correct option is (C).

   Nuclear fission and fusion are processes where the nucleus of an atom changes, releasing enormous amounts of energy. To understand why option (C) is correct, we need to recall what each of Einstein’s equations describes and which one directly explains the energy release in nuclear reactions.

   1. Identify the core phenomenon

      In both fission (splitting a heavy nucleus) and fusion (combining light nuclei), the total mass of the products is slightly less than the total mass of the reactants. This “missing” mass is called the mass defect. The energy released comes from this mass defect.

   2. Match the phenomenon to the correct equation

      Einstein’s mass-energy equation is $E = mc^2$. It states that mass and energy are interchangeable; a small amount of mass can be converted into a huge amount of energy (because $c^2$ is enormous). This directly explains why the mass defect in nuclear reactions yields the observed energy output.

   3. Eliminate the other options

      • (A) Einstein’s theory of relativity is a broader framework (including special and general relativity). While $E=mc^2$ is part of it, the question asks for the specific equation that explains the energy release, not the entire theory.
      • (B) Einstein’s specific heat equation deals with how the heat capacity of solids varies with temperature (quantum theory of solids). It has nothing to do with nuclear reactions.
      • (D) Einstein’s photoelectric equation explains the emission of electrons from a metal surface when light shines on it ($E = hf - \phi$). It is about light and electrons, not nuclear mass-energy conversion.

   4. Confirm with a classic example …

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10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If S is the surface area of a nucleus of mass number A, then (A) $S \propto A$ (B) $S \propto A^{1/3}$ (C) $S \propto A^2$ (D) $S \propto A^{2/3}$

    ›Reveal solutionSolution

    The surface area of a nucleus scales with the two-thirds power of its mass number because nuclear volume is proportional to A and the radius scales as A^{1/3}, making surface area proportional to (radius)² ∝ A^{2/3}.

    The key idea is that a nucleus is roughly spherical, and its volume is proportional to the number of nucleons (mass number A). Since volume scales as the cube of the radius, the radius scales as A^{1/3}. Surface area, being proportional to the square of the radius, therefore scales as (A^{1/3})² = A^{2/3}.

    Let’s walk through the reasoning step by step.

    1. Nuclear volume is proportional to mass number. A nucleus contains A nucleons (protons and neutrons). Since nucleons are nearly identical in size and the nucleus is densely packed, the total volume V is directly proportional to A:

    $V \propto A.$

    2. For a sphere, volume relates to radius as V ∝ R³. Assuming the nucleus is spherical (a very good approximation), the volume is

    $V = \frac{4}{3}\pi R^3 \quad \Rightarrow \quad R^3 \propto V.$

    Combining with step 1 gives

    $R^3 \propto A \quad \Rightarrow \quad R \propto A^{1/3}.$

    3. Surface area of a sphere is S = 4πR². Since S ∝ R², substituting the relation from step 2 yields

    $S \propto (A^{1/3})^2 = A^{2/3}.$

    4. Check the options.
       • (A) S ∝ A would mean surface area grows linearly with volume — impossible for a sphere (that would require a constant radius). …

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11. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.The ratio of the relative strengths of strong and weak nuclear forces is (A) $10^{13}$ (B) $10^{26}$ (C) $10^{39}$ (D) $10^{11}$

    ›Reveal solutionSolution

    The relative strengths of the strong and weak nuclear forces are compared by their coupling constants at typical energy scales; the strong force is about $10^{13}$ times stronger than the weak force, so the correct answer is (A).

    The key idea here is that fundamental forces have intrinsic "strengths" given by dimensionless coupling constants. For the strong nuclear force, the coupling constant $\alpha_s \approx 1$ at low energies, while for the weak nuclear force, the effective coupling constant is $\alpha_w \approx 10^{-5}$ (or, more precisely, the Fermi constant gives a strength ratio of about $10^{13}$). This huge difference explains why the strong force dominates inside the nucleus, while the weak force only manifests in rare processes like beta decay.

    Let’s work through the reasoning step by step:

    1. Understand what "relative strength" means.

       In particle physics, the strength of a force is quantified by its coupling constant — a dimensionless number that determines the probability of interaction. For the strong force, the coupling $\alpha_s$ is roughly $1$ at the scale of a few femtometers (the size of a nucleus). For the weak force, the effective coupling at low energies is given by the Fermi constant $G_F$ times the energy squared, but a more direct comparison uses the weak coupling constant $\alpha_w \approx 1/30$ at the mass of the W boson. However, because the weak force is mediated by very heavy particles (W and Z bosons), its effective strength at nuclear distances is much smaller.

    2. Compare the forces at typical nuclear energies.

       The strong force’s coupling is $\alpha_s \approx 1$. The weak force’s effective coupling at low energies is about $10^{-5}$ (this comes from $\alpha_w \approx 1/30$ multiplied by a factor $(m_p / m_W)^2 \approx (1/80)^2 \approx 1.6 \times 10^{-4}$, giving roughly $10^{-5}$). The ratio of strong to weak is then $1 / 10^{-5} = 10^5$. But wait — this is not the number in the options. The standard textbook comparison uses a different convention: the weak force’s strength is often quoted as $10^{-13}$ relative to the strong force, meaning the strong force is $10^{13}$ times stronger.

    3. Why $10^{13}$ and not $10^5$? …

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12. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
    Q.The ratio of the radii of the nuclei $_{13}\mathrm{Al}^{27}$ and $_{52}\mathrm{Te}^{125}$ is (A) $3 : 5$ (B) $27 : 125$ (C) $13 : 52$ (D) $14 : 73$

    ›Reveal solutionSolution

    The nuclear radius depends on the mass number as $R = R_0 A^{1/3}$, so the ratio of radii is the cube root of the ratio of mass numbers. For $^{27}\mathrm{Al}$ and $^{125}\mathrm{Te}$, this gives $3 : 5$, which is option (A).

    The key concept here is the nuclear radius formula. Unlike atomic radii, which vary in a complicated way across the periodic table, nuclear radii follow a remarkably simple rule: the volume of a nucleus is proportional to the number of nucleons (protons + neutrons). Since the nucleus is roughly spherical, its radius $R$ is proportional to $A^{1/3}$, where $A$ is the mass number. This is one of the earliest and most important empirical results in nuclear physics.

    Why does this work? Because nucleons are packed together at nearly constant density — like marbles in a bag. Double the number of marbles, you roughly double the volume, so the radius increases by the cube root of 2. So when comparing two nuclei, we don't compare their mass numbers directly; we compare the cube roots of those numbers.

    Let’s work through it step by step.

    1. Write the nuclear radius formula.

       For any nucleus, $R = R_0 A^{1/3}$, where $R_0$ is a constant (about $1.2 \times 10^{-15}$ m). The constant cancels out when taking a ratio.

    2. Identify the mass numbers.

       For $_{13}\mathrm{Al}^{27}$, the mass number $A_1 = 27$.

       For $_{52}\mathrm{Te}^{125}$, the mass number $A_2 = 125$.

    3. Set up the ratio of radii.

       $\frac{R_1}{R_2} = \frac{R_0 (27)^{1/3}}{R_0 (125)^{1/3}} = \left( \frac{27}{125} \right)^{1/3}$ …

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