Q.$^{3}_{2}\text{He}$ and $^{3}_{1}\text{H}$ nuclei have the same mass number. Do they have the same binding energy?

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The ratio of the energy released when $2.5 \times 10^{21}$ atoms of uranium undergo nuclear fission and energy equivalent of 2 mg mass of uranium is nearly (Average energy released per fission of uranium nucleus = 200 MeV) (A) $4 : 5$ (B) $8 : 9$ (C) $4 : 9$ (D) $2 : 3$

   ›Reveal solutionSolution

   Fission energy $\approx 8.0\times10^{10}\ \mathrm{J}$ versus mass-equivalent energy $mc^2=1.8\times10^{11}\ \mathrm{J}$, giving a ratio $\approx 4:9$.

   Energy from fission of $N=2.5\times10^{21}$ atoms at $200\ \mathrm{MeV}$ each:

   $E_1 = N\times200\ \mathrm{MeV}=2.5\times10^{21}\times200\times1.6\times10^{-13}\ \mathrm{J}=8.0\times10^{10}\ \mathrm{J}.$

   Energy equivalent of $2\ \mathrm{mg}$ of mass via $E=mc^2$: …

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2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.During the disintegration of a radioactive nucleus of mass number 208 at rest, two alpha particles each with kinetic energy E are emitted. The total kinetic energy of the emitted alpha particles and the daughter nucleus after the disintegration is (A) $\dfrac{51E}{25}$ (B) $\dfrac{51E}{50}$ (C) $\dfrac{52E}{25}$ (D) $\dfrac{26E}{25}$

   ›Reveal solutionSolution

   The problem involves conservation of momentum and energy in a nuclear decay where two alpha particles are emitted from a nucleus at rest. The total kinetic energy released is the sum of the kinetic energies of the two alphas and the recoiling daughter nucleus. Using momentum conservation, the daughter’s kinetic energy is found to be a fraction of the alphas’ kinetic energy, leading to the total being $\frac{52E}{25}$, so the correct option is (C).

   ---

   Concept and Intuition

   When a stationary nucleus decays, the total momentum before decay is zero. After decay, the emitted particles and the daughter nucleus must have momenta that sum to zero. Here, two alpha particles are emitted, each with the same kinetic energy $E$. They are likely emitted in opposite directions (to conserve momentum if the daughter is also moving), but the key is that the daughter nucleus recoils to balance the total momentum. The total kinetic energy released in the decay is the sum of the kinetic energies of all three particles: the two alphas and the daughter. We are given the kinetic energy of each alpha, but not the daughter’s. We must find the daughter’s kinetic energy using momentum conservation and the relation between kinetic energy and momentum.

   ---

   Step-by-step solution

   1. Set up the masses and momenta

      The parent nucleus has mass number 208 and is at rest. It decays into a daughter nucleus and two alpha particles. Each alpha particle has mass number 4, so the daughter’s mass number is $208 - 2 \times 4 = 200$.

      Let $m_\alpha$ be the mass of one alpha particle, and $m_d = 50\,m_\alpha$ be the mass of the daughter (since $200/4 = 50$).

      Each alpha has kinetic energy $E$, so its momentum magnitude is $p_\alpha = \sqrt{2 m_\alpha E}$.

   2. Apply conservation of momentum

      Since the parent is at rest, the total momentum after decay is zero. The two alpha particles are likely emitted in opposite directions (or at least with equal and opposite momenta along some axis) so that their net momentum is zero. Then the daughter nucleus must also have zero momentum — but that would mean it is at rest, which is impossible because energy would then be only $2E$, not matching any option. …

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3. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.The energy equivalent of $3.2\,\mu\text{g}$ of mass is (A) $18 \times 10^{26}\ \text{J}$ (B) $18 \times 10^{20}\ \text{MeV}$ (C) $18 \times 10^{23}\ \text{MeV}$ (D) $32 \times 10^{26}\ \text{J}$

   ›Reveal solutionSolution

   The key idea is Einstein’s mass-energy equivalence $E = mc^2$. Converting $3.2\,\mu\text{g}$ to kg and using $c = 3 \times 10^8\ \text{m/s}$ gives $2.88 \times 10^8\ \text{J}$, which matches none of the Joule options directly — but converting to MeV shows the correct match is $18 \times 10^{20}\ \text{MeV}$, option (B).

   The problem tests your ability to apply $E = mc^2$ and handle unit conversions — especially between Joules and MeV, a common trick in exam questions. The mass is given in micrograms, a tiny unit, so the energy will be large but not astronomically so. Let’s work through it carefully.

   1. Convert mass to SI units. The mass is $3.2\,\mu\text{g}$. Since $1\,\mu\text{g} = 10^{-9}\ \text{kg}$, we have

   $m = 3.2 \times 10^{-9}\ \text{kg}.$

   2. Apply $E = mc^2$. Take $c = 3 \times 10^8\ \text{m/s}$. Then

   $E = (3.2 \times 10^{-9}) \times (3 \times 10^8)^2 = 3.2 \times 10^{-9} \times 9 \times 10^{16}.$

   Multiply: $3.2 \times 9 = 28.8$, and $10^{-9} \times 10^{16} = 10^{7}$, so

   $E = 28.8 \times 10^{7} = 2.88 \times 10^{8}\ \text{J}.$

   This is $288$ million Joules — a substantial energy from a tiny mass, as relativity predicts.

   3. Check the given options.

      Option (A) is $18 \times 10^{26}\ \text{J}$ — far too large. Option (D) is $32 \times 10^{26}\ \text{J}$ — also enormous. So neither Joule option matches our result. The remaining options are in MeV, so we must convert.

   4. Convert Joules to MeV.

      Recall the conversion: $1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$. Therefore,

   $E = \frac{2.88 \times 10^{8}}{1.6 \times 10^{-13}}\ \text{MeV} = \frac{2.88}{1.6} \times 10^{8+13} = 1.8 \times 10^{21}\ \text{MeV}.$

   That’s $1.8 \times 10^{21}\ \text{MeV}$, which can be written as $18 \times 10^{20}\ \text{MeV}$.

   5. Match with options. …

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4. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The half lives of two radioactive materials A and B are respectively T and 2T. If the ratio of the initial masses of the materials A and B is 8:1, then the time after which the ratio of the masses of the materials A and B becomes 4:1 is (A) $2T$ (B) $T$ (C) $4T$ (D) $8T$

   ›Reveal solutionSolution

   Using $N = N_0 (1/2)^{t/T_{1/2}}$ for each material, the mass ratio $\frac{m_A}{m_B} = 8\left(\tfrac{1}{2}\right)^{t/2T}$; setting it to $4$ gives $t = 2T$ — option (A).

   Step-by-step solution

   Radioactive decay: $m = m_0 \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$.

   For material A (half-life $T$, initial mass $8$) and B (half-life $2T$, initial mass $1$):

   $m_A = 8\left(\frac{1}{2}\right)^{t/T},\qquad m_B = 1\cdot\left(\frac{1}{2}\right)^{t/2T}.$

   Form the ratio: …

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5. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The operation of a nuclear reactor is said to be critical when the value of neutron multiplication factor K is (A) $K = 0$ (B) $K > 1$ (C) $K = 1$ (D) $0 < K < 1$

   ›Reveal solutionSolution

   The neutron multiplication factor $K$ measures whether a nuclear chain reaction is self-sustaining. Criticality occurs when exactly one neutron from each fission goes on to cause another fission, so $K = 1$. The correct option is (C).

   The key idea is that a nuclear reactor’s operation depends on a chain reaction: each fission event releases neutrons, and some of those neutrons cause further fissions. The multiplication factor $K$ is the average number of neutrons from one fission that successfully produce another fission.

   • If $K < 1$, the reaction dies out (subcritical).
   • If $K > 1$, the reaction grows exponentially (supercritical) — dangerous if uncontrolled.
   • If $K = 1$, the reaction is exactly self-sustaining (critical), which is the steady, controlled state for power generation.

   Thus, “critical” means the chain reaction is balanced: each fission leads to exactly one more fission on average.

   1. Define $K$: The neutron multiplication factor is the ratio of neutrons in one generation to the number in the previous generation. In a steady reactor, we want this ratio to be 1 so that the neutron population remains constant.

   2. Interpret the options:

      • $K = 0$: No chain reaction at all (impossible for a working reactor).
      • $K > 1$: Exponential increase — used for startup or weapons, not steady operation.
      • $K = 1$: Perfect balance — the reactor is critical.
      • $0 < K < 1$: The reaction fades away — subcritical. …

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.An $\alpha$-particle of energy ‘E’ is liberated during the decay of a nucleus of mass number 236. The total energy released in this process is (A) 58 E (B) 59 E (C) $\dfrac{58E}{59}$ (D) $\dfrac{59E}{58}$

   ›Reveal solutionSolution

   The key idea is that the total energy released equals the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus, which share the decay energy in inverse proportion to their masses. Using conservation of momentum and the given mass numbers, the total energy is $\frac{59}{58}E$, so the correct option is (D).

   When a nucleus decays by emitting an alpha particle, the energy released (the Q-value of the decay) is shared between the alpha particle and the recoiling daughter nucleus. The alpha particle gets most of the energy because it is much lighter, but the daughter nucleus also carries away some kinetic energy to conserve momentum. The problem gives the alpha particle’s kinetic energy as $E$ and asks for the total energy released. The trick is to relate the masses (via mass numbers) to the energy split.

   Why this approach works:

   In any decay at rest, momentum is conserved. Since the parent nucleus is initially stationary, the alpha particle and the daughter nucleus must have equal and opposite momenta. Kinetic energy is $p^2/(2m)$, so for a given momentum, the energy is inversely proportional to mass. Thus, the ratio of the alpha’s energy to the daughter’s energy is the inverse ratio of their masses. The total energy is then the sum, which we can express in terms of the given $E$.

   Step-by-step reasoning:

   1. Identify the masses involved.

      The parent nucleus has mass number 236. It decays into an alpha particle (mass number 4) and a daughter nucleus. The daughter’s mass number is $236 - 4 = 232$.

      Let $m_\alpha = 4u$ and $m_d = 232u$, where $u$ is the atomic mass unit. (We use mass numbers as proportional to masses; the approximation is excellent for this problem.)

   2. Apply conservation of momentum.

      Initially, the parent is at rest, so total momentum is zero. After decay, the alpha and daughter move in opposite directions with momenta of equal magnitude:

   $p_\alpha = p_d = p.$

   3. Relate kinetic energy to momentum. Kinetic energy for a particle is $K = \frac{p^2}{2m}$. Therefore:

   $K_\alpha = \frac{p^2}{2m_\alpha}, \quad K_d = \frac{p^2}{2m_d}.$

   The ratio of their kinetic energies is:

   $\frac{K_\alpha}{K_d} = \frac{m_d}{m_\alpha} = \frac{232}{4} = 58.$ …

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7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If the ratio of the radii of nuclei $_{52}^XA$ and $_{13}^{27}Al$ is $5:3$, then the number of neutrons in the nucleus X is (A) $52$ (B) $63$ (C) $27$ (D) $73$

   ›Reveal solutionSolution

   The nuclear radius scales as $R \propto A^{1/3}$, so the ratio of radii gives the ratio of mass numbers; solving yields $A_X = 125$, and with atomic number $Z=52$, neutrons $N = A - Z = 73$.

   The key idea is that nuclear radii follow a simple empirical law: $R = R_0 A^{1/3}$, where $A$ is the mass number (total protons + neutrons). This means the volume of a nucleus is proportional to $A$, so comparing radii directly gives a comparison of mass numbers. Once we find the unknown mass number, subtracting the given atomic number (52) gives the number of neutrons.

   Let’s work through it step by step.

   1. Write the radius formula for both nuclei. For any nucleus, $R = R_0 A^{1/3}$, where $R_0$ is a constant (about $1.2 \times 10^{-15}$ m, but it cancels). For the unknown nucleus $_{52}^X A$ (let’s call its mass number $A_X$):

   $R_X = R_0 (A_X)^{1/3}$

   For $_{13}^{27}\text{Al}$:

   $R_{\text{Al}} = R_0 (27)^{1/3}$

   2. Set up the given ratio. The problem states:

   $\frac{R_X}{R_{\text{Al}}} = \frac{5}{3}$

   Substitute the expressions:

   $\frac{R_0 (A_X)^{1/3}}{R_0 (27)^{1/3}} = \frac{5}{3}$

   The $R_0$ cancels, leaving:

   $\frac{(A_X)^{1/3}}{(27)^{1/3}} = \frac{5}{3}$

   3. Simplify the denominator. Since $27 = 3^3$, we have $(27)^{1/3} = 3$. So:

   $\frac{(A_X)^{1/3}}{3} = \frac{5}{3}$

   Multiply both sides by 3:

   $(A_X)^{1/3} = 5$

   4. Solve for $A_X$. Cube both sides: $A_X = 5^3 = 125$ …

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8. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.Half-life periods of two nuclei A and B are T and 2T respectively. Initially A and B have same number of nuclei. After a time of 4T, the ratio of the remaining number of nuclei of A and B is (A) 1 : 16 (B) 1 : 4 (C) 1 : 1 (D) 1 : 2

   ›Reveal solutionSolution

   The ratio of remaining nuclei after time $4T$ is found by applying the exponential decay law to each nucleus, using their respective half-lives. The result is $1 : 4$, so option (B) is correct.

   The key idea is that radioactive decay follows an exponential law: the number of nuclei remaining after time $t$ is $N = N_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$, where $T_{1/2}$ is the half-life. The ratio of two samples depends only on how many half-lives each has experienced.

   1. Identify the number of half-lives for each nucleus. For nucleus A, half-life $T_A = T$. Time elapsed is $4T$, so the number of half-lives is

   $n_A = \frac{4T}{T} = 4.$

   For nucleus B, half-life $T_B = 2T$. The number of half-lives is

   $n_B = \frac{4T}{2T} = 2.$

   2. Write the remaining number for each. Let the initial number of each be $N_0$. After $n$ half-lives,

   $N_A = N_0 \left( \frac{1}{2} \right)^{4} = \frac{N_0}{16},$

   $N_B = N_0 \left( \frac{1}{2} \right)^{2} = \frac{N_0}{4}.$

   3. Compute the ratio $N_A : N_B$.

   $\frac{N_A}{N_B} = \frac{N_0/16}{N_0/4} = \frac{1}{16} \times \frac{4}{1} = \frac{1}{4}.$

   So the ratio is $1 : 4$. …

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9. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $F_1$ and $F_2$ are the relative strengths of the gravitational and weak nuclear forces respectively, then $\dfrac{F_2}{F_1}$ is nearly (A) $100$ (B) $10^{39}$ (C) $10^{13}$ (D) $10^{26}$

   ›Reveal solutionSolution

   The ratio of the weak nuclear force to the gravitational force is enormous — about $10^{33}$ — but the given options are based on the relative strengths of the forces compared to the strong force, so the correct choice is (D) $10^{26}$.

   The key idea: In physics, "relative strengths" of fundamental forces are usually quoted as dimensionless numbers compared to the strong nuclear force (taken as 1). Gravitational force is about $10^{-39}$ times the strong force, and the weak nuclear force is about $10^{-13}$ times the strong force. Their ratio $\frac{F_2}{F_1}$ is then $\frac{10^{-13}}{10^{-39}} = 10^{26}$.

   ---

   Why this approach works

   The question doesn't give absolute numbers — it asks for the ratio of two forces. That means we only need their relative sizes on a common scale. The standard scale in particle physics compares all forces to the strong nuclear force. Memorizing these approximate exponents is the direct path.

   ---

   Step-by-step reasoning

   1. Recall the approximate relative strengths of the four fundamental forces (strong nuclear = 1):

      • Strong nuclear: $1$
      • Electromagnetic: $\approx 10^{-2}$
      • Weak nuclear: $\approx 10^{-13}$
      • Gravitational: $\approx 10^{-39}$

   2. Identify $F_1$ and $F_2$ from the problem:

      • $F_1$ = relative strength of gravitational force = $10^{-39}$
      • $F_2$ = relative strength of weak nuclear force = $10^{-13}$

   3. Compute the ratio:

      $$$$

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10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.In the following nuclear reaction X is $^{13}_{27}\mathrm{Al} + ^{4}_{2}\mathrm{He} \rightarrow ^{1}_{0}n + X$ (A) $^{31}_{15}\mathrm{P}$ (B) $^{30}_{14}\mathrm{Si}$ (C) $^{30}_{15}\mathrm{P}$ (D) $^{31}_{15}\mathrm{Si}$

    ›Reveal solutionSolution

    In a nuclear reaction, both mass number and atomic number must be conserved.

    For $^{27}_{13}\mathrm{Al} + ^{4}_{2}\mathrm{He} \rightarrow ^{1}_{0}n + X$, the missing nucleus X has mass number $30$ and atomic number $15$, which is $^{30}_{15}\mathrm{P}$ — option (C).

    The key idea is conservation of nucleon number and charge in nuclear reactions. Just like in a chemical equation, the total "mass" (mass number) and total "charge" (atomic number) on the left must equal those on the right. Here, we know the neutron has mass number 1 and atomic number 0, so we can solve for X by simple subtraction.

    1. Write down the known numbers

       Left side:

       • Aluminium: mass number = 27, atomic number = 13
       • Helium (alpha particle): mass number = 4, atomic number = 2 Total left: mass = $27 + 4 = 31$, atomic = $13 + 2 = 15$

    2. Right side so far

       • Neutron: mass number = 1, atomic number = 0 So the missing nucleus X must supply the rest: Mass number of X = $31 - 1 = 30$ Atomic number of X = $15 - 0 = 15$

    3. Identify the element

       Atomic number 15 is phosphorus (P). So X is $^{30}_{15}\mathrm{P}$.

    4. Check the options

       (A) $^{31}_{15}\mathrm{P}$ — wrong mass number …

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11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.In the nuclear fission of one nucleus of U$^{235}$ the energy released is 188 MeV. The energy released in the nuclear fission of 235 g of U$^{235}$ is nearly (Avogadro number = $6.02 \times 10^{23}$ mol$^{-1}$) (A) $28.8 \times 10^{12}$ J (B) $23.5 \times 10^{12}$ J (C) $36.2 \times 10^{12}$ J (D) $18.11 \times 10^{12}$ J

    ›Reveal solutionSolution

    The problem asks for the total energy released when 235 g of U‑235 undergoes fission, given 188 MeV per nucleus. The key is to find the number of nuclei in 235 g using Avogadro’s number, multiply by the energy per fission, and convert MeV to joules. The result is about $1.81 \times 10^{13}$ J, which matches option (D).

    The core idea is simple: energy per fission × number of fissions = total energy.

    Since 235 g of U‑235 is exactly one mole (because the atomic mass is 235 u), the number of nuclei is Avogadro’s number. Then we just need to convert the energy from MeV to joules (1 MeV = $1.602 \times 10^{-13}$ J).

    1. Find the number of nuclei in 235 g of U‑235

       The atomic mass of U‑235 is 235 g/mol, so 235 g is exactly 1 mole.

       Number of nuclei = Avogadro’s number = $6.02 \times 10^{23}$.

    2. Energy released per nucleus

       Given: 188 MeV per fission.

    3. Convert MeV to joules

       $1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}$

       So $188 \text{ MeV} = 188 \times 1.602 \times 10^{-13} \text{ J}$.

    4. Calculate total energy

       Total energy = (number of nuclei) × (energy per nucleus in J)

    $E = (6.02 \times 10^{23}) \times (188 \times 1.602 \times 10^{-13})$

    First compute the product inside:

    $188 \times 1.602 = 301.176$

    So $188 \times 1.602 \times 10^{-13} = 3.01176 \times 10^{-11}$ J per nucleus.

    Then multiply:

    $E = 6.02 \times 10^{23} \times 3.01176 \times 10^{-11} = (6.02 \times 3.01176) \times 10^{12}$

    $6.02 \times 3.01176 \approx 18.13$

    So $E \approx 1.813 \times 10^{13}$ J.

    5. Match with options …

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12. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.An electron and proton are produced by breaking of a neutron. If the mass of neutron, electron and proton are $1.675 \times 10^{-27}$ kg, $9 \times 10^{-31}$ kg and $1.6730 \times 10^{-27}$ kg respectively, find the amount of energy released in this process. (consider velocity of light as $3 \times 10^8$ m/s) (A) 0.66 MeV (B) 0.59 MeV (C) 0.62 MeV (D) 0.68 MeV

    ›Reveal solutionSolution

    The neutron decays into a proton and an electron, and the tiny missing mass is converted into kinetic energy. Using $E = \Delta m c^2$, the energy released comes out to 0.62 MeV, which matches option (C).

    The process described is beta-minus decay of a free neutron:

    $n \rightarrow p + e^- + \bar{\nu}_e$ (the antineutrino is nearly massless and carries negligible energy here). The total mass of the products is slightly less than the mass of the neutron. That missing mass — the mass defect — is converted entirely into kinetic energy shared by the proton, electron, and antineutrino. The question asks for the total energy released, which is simply $\Delta m c^2$.

    The key idea: mass is a form of energy. When a system loses mass, that mass reappears as kinetic energy (or radiation). Here, we don't need to track individual particle energies — just the mass difference.

    1. Write the mass defect Mass of neutron: $m_n = 1.675 \times 10^{-27}\ \text{kg}$ Mass of proton: $m_p = 1.6730 \times 10^{-27}\ \text{kg}$ Mass of electron: $m_e = 9 \times 10^{-31}\ \text{kg}$ The mass of the products is $m_p + m_e$. So the mass defect is:

    $\Delta m = m_n - (m_p + m_e)$

    2. Calculate the numerical value First add the product masses:

    $m_p + m_e = 1.6730 \times 10^{-27} + 0.0009 \times 10^{-27} = 1.6739 \times 10^{-27}\ \text{kg}$

    (since $9 \times 10^{-31} = 0.0009 \times 10^{-27}$).

    Then:

    $\Delta m = (1.6750 - 1.6739) \times 10^{-27} = 0.0011 \times 10^{-27}\ \text{kg}$

    That is:

    $\Delta m = 1.1 \times 10^{-30}\ \text{kg}$

    3. Apply Einstein’s mass-energy equivalence

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

    with $c = 3 \times 10^8\ \text{m/s}$. So:

    $E = (1.1 \times 10^{-30}) \times (3 \times 10^8)^2$

    $E = 1.1 \times 10^{-30} \times 9 \times 10^{16}$

    $E = 9.9 \times 10^{-14}\ \text{J}$

    4. Convert joules to MeV …

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