Q.The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei $^{41}_{20}\text{Ca}$ and $^{27}_{13}\text{Al}$ from the following data:
$m(^{40}_{20}\text{Ca}) = 39.962591$ u
$m(^{41}_{20}\text{Ca}) = 40.962278$ u
$m(^{26}_{13}\text{Al}) = 25.986895$ u
$m(^{27}_{13}\text{Al}) = 26.981541$ u

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.What is the energy released in the following fission reaction? $^{236}_{z}U \rightarrow {}^{117}_{a}X + {}^{117}_{(z-a)}Y + 2\left({}^{1}_{0}n\right)$ Given that the binding energy per nucleon of X and Y is 8.8 MeV and that of $^{236}_{z}U$ is 7.5 MeV (A) 256.4 MeV (B) 248.6 MeV (C) 274.2 MeV (D) 289.2 MeV

   ›Reveal solutionSolution

   Energy released in fission equals the increase in total nuclear binding energy from reactant to products; computing this gives 289.2 MeV.

   Concept and Intuition

   Nuclear fission releases energy because the fragment nuclei are, per nucleon, more tightly bound than the original heavy nucleus (binding energy per nucleon peaks around mass number ~56 and is higher for the mid-mass fragments than for very heavy nuclei like uranium). The energy released is exactly the difference between the total binding energy of the products and that of the reactant — free neutrons contribute zero binding energy since they are not bound in any nucleus.

   Step-by-Step Solution

   1. Total binding energy of the reactant, $^{236}_ZU$: $BE_{reactant} = 236 \times 7.5 = 1770$ MeV.
   2. The two product nuclei X and Y together have $117+117=234$ nucleons, each bound at 8.8 MeV/nucleon: $BE_{products} = 234\times8.8 = 2059.2$ MeV.
   3. The 2 free neutrons emitted carry no binding energy (they are not part of a nucleus).
   4. Energy released $=BE_{products}-BE_{reactant} = 2059.2-1770 = 289.2$ MeV.

   Common Mistakes

   • Including the 2 free neutrons' mass number (238 total) in the binding-energy-per-nucleon multiplication for the products, instead of only the 234 nucleons that are actually bound in X and Y.
   • Subtracting in the wrong order (reactant − products) and getting a negative/wrong-magnitude answer.

   ✓Final answer

   The correct option is (D) — 289.2 MeV.

   ANSWER: D

2. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Energy released in the fission of a single $_{92}U^{235}$ nucleus is 200 MeV. The fission rate of $_{92}U^{235}$ fueled reactor operating at a power level of 5W is (A) $1.56\times10^{10}\ s^{-1}$ (B) $1.56\times10^{11}\ s^{-1}$ (C) $1.56\times10^{16}\ s^{-1}$ (D) $1.56\times10^{-17}\ s^{-1}$

   ›Reveal solutionSolution

   Divide reactor power by energy released per fission event to get the fission rate: $\approx 1.56\times10^{11}$ fissions per second.

   Concept and Intuition

   A nuclear reactor's power output is the total energy released per second, which equals (energy released per fission) × (number of fissions per second). So the fission rate is just power divided by per-fission energy, after converting MeV to joules using $1\ MeV = 1.6\times10^{-13}\ J$.

   Step-by-Step Solution

   1. Convert energy per fission to SI: $E = 200\ MeV = 200 \times 1.6\times10^{-13}\ J = 3.2\times10^{-11}\ J$.
   2. Reactor power $P = 5\ W = 5\ J/s$.
   3. Fission rate $R = \dfrac{P}{E} = \dfrac{5}{3.2\times10^{-11}} = 1.5625\times10^{11}$ fissions per second.

   Common Mistakes

   • Using $1\ MeV = 1.6\times10^{-19}\ J$ (that's the eV-to-J conversion, not MeV) — this is off by a factor of $10^6$ and lands on a wrong-magnitude option like (D).
   • Arithmetic slip when dividing by a number in scientific notation, landing on $10^{10}$ or $10^{16}$ instead of $10^{11}$.

   ✓Final answer

   The correct option is (B) — $1.56\times10^{11}\ s^{-1}$.

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Each nuclear fission of $^{235}\mathrm{U}$ releases 200 MeV of energy. If a reactor generates 1 MW power, then the rate of fission in the reactor is (A) $3.125\times10^6$ (B) $3.125\times10^8$ (C) $3.125\times10^{10}$ (D) $3.125\times10^{16}$

   ›Reveal solutionSolution

   This tests converting a reactor's power output into a fission rate using the known energy released per fission event. The rate comes out to $3.125\times10^{16}$ fissions/second.

   Concept and Intuition

   Power is energy delivered per unit time. If each fission event reliably releases a fixed amount of energy, then the number of fission events needed per second to sustain a given power output is simply that power divided by the energy released per event — exactly like asking how many coins of a given value you need per second to make up a certain rate of payment.

   Step-by-Step Solution

   1. Energy released per fission: $200\ \text{MeV}$. Convert to joules using $1\ \text{eV}=1.6\times10^{-19}$ J:

   $E_{\text{fission}}=200\times10^6\times1.6\times10^{-19}\ \text{J}=3.2\times10^{-11}\ \text{J}$

   2. Reactor power: $P=1\ \text{MW}=10^6\ \text{W}=10^6\ \text{J/s}$.
   3. Rate of fission (fissions per second):

   $n=\frac{P}{E_{\text{fission}}}=\frac{10^6}{3.2\times10^{-11}}=3.125\times10^{16}\ \text{s}^{-1}$

   Common Mistakes

   • Forgetting to convert MeV to joules (leaving the huge unit mismatch uncorrected).
   • Arithmetic slip in the division — it helps to write $10^6/3.2\times10^{-11}=(1/3.2)\times10^{17}=0.3125\times10^{17}=3.125\times10^{16}$.

   ✓Final answer

   The correct option is (D) — $3.125\times10^{16}$.

   ANSWER: D

4. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.The energy released in the fission of one ${}_{92}\text{U}^{235}$ nucleus is 200 MeV. The energy released in the fission of 235 g mass of ${}_{92}\text{U}^{235}$ is nearly (A) $15.84 \times 10^{12}$ J (B) $19.27 \times 10^{12}$ J (C) $13.59 \times 10^{12}$ J (D) $17.73 \times 10^{12}$ J

   ›Reveal solutionSolution

   235 g of U-235 is exactly one mole of nuclei; multiplying the per-nucleus fission energy by Avogadro's number gives $19.27\times10^{12}$ J.

   Concept and Intuition

   The atomic mass of U-235 in grams equals one mole of that isotope (this is precisely why 235 g is chosen in the problem — it's a clean way to get exactly $N_A$ nuclei). So the total fission energy from 235 g is just the per-nucleus energy times Avogadro's number.

   Step-by-Step Solution

   1. Convert per-nucleus energy to joules: $200\ \text{MeV} = 200\times10^6\times1.6\times10^{-19}\ \text{J} = 3.2\times10^{-11}$ J.
   2. Number of nuclei in 235 g: $N=N_A=6.022\times10^{23}$.
   3. Total energy: $E = 3.2\times10^{-11}\times6.022\times10^{23}$.
   4. $3.2\times6.022=19.27$, and the powers of ten give $10^{-11+23}=10^{12}$, so $E\approx19.27\times10^{12}$ J.

   Common Mistakes

   • Using the wrong conversion for MeV to Joules (should be $1.6\times10^{-13}$ J per MeV, i.e. $1.6\times10^{-19}\times10^6$).
   • Forgetting that 235 g corresponds to exactly one mole (i.e., $N_A$ atoms), and instead trying to use some other mass-to-number conversion.

   ✓Final answer

   The correct option is (B) — $19.27\times10^{12}$ J.

   ANSWER: B

5. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The decrease in each day in the Uranium mass of the material in a Uranium reactor operating at a power of 12 MW is (Energy released in one $_{92}\mathrm{U}^{235}$ fission is about 200 MeV) (A) $12.64 \times 10^{-2}$ kg (B) $11.50 \times 10^{-2}$ g (C) 12.64 kg (D) 12.64 g

   ›Reveal solutionSolution

   This tests converting reactor power output into a mass of fissile fuel consumed per day, via the energy released per fission event.

   Concept and Intuition

   A nuclear reactor's power output comes from a huge number of individual fission events, each releasing a fixed amount of energy (~200 MeV for U-235). Knowing the total energy needed per day and the energy per fission tells us how many nuclei fission per day; multiplying by the mass per nucleus (via Avogadro's number) gives the mass consumed.

   Step-by-Step Solution

   1. Total energy released per day: $E=P\times t=12\times10^6\ \text{W}\times(24\times3600\ \text{s})=12\times10^6\times86400=1.0368\times10^{12}$ J.
   2. Energy released per fission: $200\ \text{MeV}=200\times10^6\times1.6\times10^{-19}\ \text{J}=3.2\times10^{-11}$ J.
   3. Number of fissions per day: $N=\dfrac{E}{E_{fission}}=\dfrac{1.0368\times10^{12}}{3.2\times10^{-11}}\approx3.24\times10^{22}$.
   4. Mass of one mole of U-235 is 235 g, containing $N_A=6.022\times10^{23}$ atoms; mass consumed $=\dfrac{N}{N_A}\times235=\dfrac{3.24\times10^{22}}{6.022\times10^{23}}\times235\approx0.0538\times235\approx12.64$ g.

   Common Mistakes

   • Forgetting to convert the day into seconds (86400 s) before computing total energy.
   • Mixing up grams and kilograms in the final answer.

   ✓Final answer

   The correct option is (D) — 12.64 g.

   ANSWER: D

6. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If the energy released per fission of a $^{235}_{92}$U nucleus is 200 MeV, the energy released in the fission of 0.1 kg of $^{235}_{92}$U in kilowatt-hour is. (A) $22.8\times10^5$ (B) $22.8\times10^7$ (C) $11.4\times10^5$ (D) $820\times10^{10}$

   ›Reveal solutionSolution

   Count the U-235 atoms in the given mass, multiply by the per-fission energy, then convert joules to kilowatt-hours.

   Concept and Intuition

   Each fission event releases a fixed amount of energy (given as 200 MeV). To get the total energy from a macroscopic mass, we need the number of atoms/nuclei present, found via moles and Avogadro's number. The final conversion from joules to kWh uses the standard $1\,\text{kWh}=3.6\times10^6$ J.

   Step-by-Step Solution

   1. Moles in 0.1 kg = 100 g: $n=\dfrac{100}{235}=0.4255$ mol.
   2. Number of atoms: $N = 0.4255\times6.022\times10^{23}=2.562\times10^{23}$.
   3. Energy per fission in joules: $200\ \text{MeV} = 200\times1.6\times10^{-13}\ \text J = 3.2\times10^{-11}$ J.
   4. Total energy: $E = 2.562\times10^{23}\times3.2\times10^{-11} = 8.20\times10^{12}$ J.
   5. Convert to kWh: $\dfrac{8.20\times10^{12}}{3.6\times10^{6}} \approx 2.278\times10^{6}\ \text{kWh} = 22.8\times10^{5}$ kWh.

   Common Mistakes

   • Using 0.1 kg as if it's grams directly in the mole calculation (must convert to 100 g first, using molar mass 235 g/mol).
   • Forgetting to convert MeV to joules before totaling, or mixing unit conversions when going to kWh.

   ✓Final answer

   The correct option is (A) — $22.8\times10^5$.

   ANSWER: A

7. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.In a nuclear reactor, the fuel is consumed at the rate of $1\times10^{-3}\ \mathrm{g\,s^{-1}}$. The power generated in kW is (A) $9\times10^{14}$ (B) $9\times10^{7}$ (C) $9\times10^{8}$ (D) $9\times10^{12}$

   ›Reveal solutionSolution

   Converting the mass-consumption rate to a power via $E=mc^2$ gives $9\times10^7$ kW.

   Concept and Intuition

   In a nuclear reactor, mass is converted into energy according to Einstein's mass-energy equivalence, $E=mc^2$. If a certain mass of fuel is "consumed" (i.e., converted to energy) every second, then the energy released per second — which is exactly the power output — follows directly by plugging the mass rate into this formula.

   Step-by-Step Solution

   1. Given mass consumption rate: $\dfrac{dm}{dt} = 1\times10^{-3}\ \text{g/s} = 1\times10^{-6}\ \text{kg/s}$.
   2. Power (energy released per second): $P = \dfrac{dm}{dt}\,c^2$.
   3. $c^2 = (3\times10^8)^2 = 9\times10^{16}\ \text{m}^2/\text{s}^2$.
   4. $P = 1\times10^{-6}\times9\times10^{16} = 9\times10^{10}\ \text{W}$.
   5. Convert to kW: $P = \dfrac{9\times10^{10}}{1000} = 9\times10^{7}\ \text{kW}$.

   Common Mistakes

   • Forgetting to convert grams to kilograms before applying $E=mc^2$ (an error of $10^3$).
   • Forgetting to convert the final answer from watts to kilowatts (another error of $10^3$), which would land on a different power-of-ten option.

   ✓Final answer

   The correct option is (B) — $9\times10^{7}$.

   ANSWER: B

8. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.If the speed of the light was half the present value, the energy released in a nuclear reaction decreases by (A) 100% (B) 75% (C) 50% (D) 25%

   ›Reveal solutionSolution

   Since nuclear energy release follows $E=mc^2$, halving $c$ reduces $c^2$ (and hence $E$) to one-quarter of its original value — a 75% decrease.

   Concept and Intuition

   The mass–energy equivalence relation $E=mc^2$ shows energy scales with the square of the speed of light, not linearly with it. So even a modest change in $c$ produces an amplified (squared) change in the energy released for the same mass defect.

   Step-by-Step Solution

   1. Original energy: $E = mc^2$.
   2. New speed: $c' = c/2$, so new energy $E' = m(c')^2 = m\dfrac{c^2}{4} = \dfrac{E}{4}$.
   3. Decrease $= E - E' = E - \dfrac{E}{4} = \dfrac{3E}{4}$.
   4. Percentage decrease $= \dfrac{3E/4}{E}\times100\% = 75\%$.

   Common Mistakes

   • Forgetting the square in $E=mc^2$ and assuming a 50% decrease (linear scaling) instead of 75% (quadratic scaling).

   ✓Final answer

   The correct option is (B) — 75%.

   ANSWER: B

9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The nucleus having highest binding energy per nucleon is (A) $^{16}_{8}\text{O}$ (B) $^{56}_{26}\text{Fe}$ (C) $^{208}_{84}\text{Pb}$ (D) $^{4}_{2}\text{He}$

   ›Reveal solutionSolution

   The binding-energy-per-nucleon curve peaks around iron (A ≈ 56); among the given nuclei, Fe-56 sits closest to that peak.

   Concept and Intuition

   The famous binding energy per nucleon vs. mass number curve rises steeply for light nuclei, peaks around $A=56$ (iron group), then falls slowly for heavier nuclei. This is why fusion releases energy for light elements and fission releases energy for heavy elements — both processes move nuclei toward the iron peak.

   Step-by-Step Solution

   1. Recall approximate binding energy per nucleon values: $^{4}_2\text{He}\approx7.1$ MeV, $^{16}_8\text{O}\approx8.0$ MeV, $^{56}_{26}\text{Fe}\approx8.8$ MeV, $^{208}_{82}\text{Pb}\approx7.9$ MeV.
   2. Comparing these, Fe-56 has the maximum value among the choices.
   3. This matches the known peak of the BE/nucleon curve at mass number ~56.

   Common Mistakes

   • Assuming binding energy per nucleon keeps increasing with mass number — it actually peaks and then decreases.
   • Confusing total binding energy (which does increase with A) with binding energy per nucleon (which peaks at iron).

   ✓Final answer

   The correct option is (B) — $^{56}_{26}\text{Fe}$.

   ANSWER: B

10. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.The process that mainly takes place in stars to produce energy (A) nuclear fission (B) nuclear fusion (C) ionization (D) annihilation

    ›Reveal solutionSolution

    This tests basic astrophysics: the source of a star's energy. Answer: nuclear fusion.

    Concept and Intuition

    Stars like our Sun are powered by thermonuclear fusion reactions in their cores, where

    extreme temperature and pressure force light nuclei (mainly hydrogen) to fuse into

    heavier nuclei (helium), converting a tiny fraction of mass into a huge amount of

    energy ($E=mc^2$) that is radiated as heat and light.

    Step-by-Step Solution

    1. Compare the options: nuclear fission is the splitting of heavy nuclei (used in reactors on Earth, not the dominant stellar process); ionization and annihilation don't release nearly enough sustained energy to power a star for billions of years.
    2. In stars, hydrogen nuclei fuse (proton-proton chain, or CNO cycle in heavier stars) to form helium, releasing energy continuously over the star's lifetime.
    3. This is nuclear fusion — the correct answer.

    Common Mistakes

    • Confusing fusion (combining light nuclei, powers stars) with fission (splitting heavy nuclei, powers nuclear reactors on Earth).

    ✓Final answer

    The correct option is (B) — nuclear fusion.

    ANSWER: B

11. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.An experimental train uses 1 g of nuclear material to run. 90 % of the produced energy is wasted to overcome the frictional force between the wheels and the track. If the weight of each coach is 18 ton (18000 kg) and it runs at a speed of 100 m s$^{-1}$, the number of coaches the engine can drag at a time is (Speed of light in vacuum is $= 3\times10^8$ m s$^{-1}$) (A) 100 (B) 1000 (C) 10000 (D) 100000

    ›Reveal solutionSolution

    Mass-energy from the fuel splits 90%-wasted / 10%-useful; the useful 10% supplies the kinetic energy of the coaches at their running speed, from which the number of coaches follows.

    Concept and Intuition

    By $E=mc^2$, even a tiny mass of nuclear fuel releases enormous energy. Here, 90% of that energy is dissipated as heat overcoming friction between wheels and track (this is the "wasted" part), leaving 10% as the useful energy that actually goes into moving the train — i.e. giving the coaches their kinetic energy at running speed.

    Step-by-Step Solution

    1. Total energy released: $E = mc^2 = (10^{-3}\ \text{kg})\times(3\times10^8\ \text{m/s})^2 = 9\times10^{13}$ J.
    2. Useful (non-wasted) energy: $10\%$ of $E$ = $9\times10^{12}$ J.
    3. This equals the total kinetic energy of $n$ coaches, each of mass $M=18000$ kg, moving at $v=100$ m/s:

    $9\times10^{12} = \frac12\, n\, M\, v^2 = \frac12\, n \times 18000 \times (100)^2.$

    4. Solve: $\frac12\times18000\times10^4 = 9\times10^7$, so $n = \dfrac{9\times10^{12}}{9\times10^7} = 10^5 = 100000$.

    Common Mistakes

    • Using the 90% "wasted" energy instead of the 10% useful energy in the kinetic-energy equation.
    • Forgetting to convert grams to kilograms in $E=mc^2$.

    ✓Final answer

    The correct option is (D) — 100000.

    ANSWER: D

12. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.Which of the following statement is correct? (A) The rest mass of the stable nucleus is less than the sum of the rest masses of its separated nucleons (B) The rest mass of the stable nucleus is greater than the sum of the rest masses of its separated nucleons (C) In nuclear fission, energy is released by fusion is of two nuclei of medium mass (approximate 100amu) (D) In nuclear fission, energy is released by fragmentation is very low atomic mass nucleus

    ›Reveal solutionSolution

    A stable nucleus's rest mass is less than the sum of its free nucleons' rest masses — the mass defect that accounts for its binding energy — making option (A) correct.

    Concept and Intuition

    Binding a set of nucleons together into a nucleus releases energy (the binding energy), and by mass-energy equivalence, that released energy corresponds to a loss of rest mass. So a bound (stable) nucleus always weighs less than its separated, free constituent nucleons — this "mass defect" is a foundational nuclear-physics fact, and it is precisely why nuclear reactions that increase binding energy per nucleon (fission of heavy nuclei, or fusion of light nuclei) release energy.

    Step-by-Step Solution

    1. Binding energy: $BE=(\text{sum of rest masses of free nucleons}-\text{rest mass of nucleus})c^2 > 0$ for any stable, bound nucleus.
    2. This directly means: rest mass of nucleus $<$ sum of rest masses of separated nucleons — confirming option (A) and ruling out (B).
    3. Nuclear fission is the splitting of a heavy nucleus (like U-235) into medium-mass fragments (~100 amu each), releasing energy because the fragments are more tightly bound per nucleon than the parent — this is fission, not "fusion of two nuclei of medium mass" as option (C) mislabels it.
    4. Nuclear fusion is the combining of very light nuclei (like hydrogen isotopes), releasing energy — option (D) mislabels this process as "fission... of very low atomic mass nucleus," swapping the terminology.

    Common Mistakes

    • Swapping "fission" and "fusion" in describing which process combines light nuclei versus splits heavy ones (options C and D deliberately test this common confusion).
    • Thinking binding makes a nucleus heavier than its parts, rather than lighter (the mass defect is easy to get backwards).

    ✓Final answer

    The correct option is (A) — The rest mass of the stable nucleus is less than the sum of the rest masses of its separated nucleons.

    ANSWER: A