Q.From the relation $R = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$).

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the nuclear force between a proton and a neutron is attractive, then the distance between them can be (A) $0.12$ fm (B) $10^{-3}$ fm (C) $1.1$ fm (D) $0.3$ fm

   ›Reveal solutionSolution

   The nuclear force is attractive only within a very narrow range (about 1–2 fm), so the only plausible distance among the options is 1.1 fm — option (C).

   The key concept here is the range of the strong nuclear force. Unlike gravity or electromagnetism, which have infinite range, the strong force that binds protons and neutrons in a nucleus acts only over extremely short distances — roughly the size of a medium-sized nucleus. If two nucleons are too close (much less than 0.5 fm), the repulsive core of the nuclear force dominates; if they are too far (beyond about 2–3 fm), the force becomes negligible. So the distance for an attractive interaction must lie in a specific window.

   Let’s examine each option:

   1. Option (A): 0.12 fm

      This is far too small. At such a short distance, nucleons would experience the repulsive core of the nuclear force (like two hard spheres pressing together). The strong force is attractive only beyond about 0.5 fm. So 0.12 fm is in the repulsive region.

   2. Option (B): $10^{-3}$ fm

      This is even smaller — a thousandth of a femtometer. At this scale, the nucleons would essentially overlap, and the force is strongly repulsive. Not attractive.

   3. Option (C): 1.1 fm

      This is right in the sweet spot. The typical separation between nucleons in a nucleus is about 1–2 fm. At 1.1 fm, the nuclear force is strongly attractive — this is the distance where the potential energy is most negative (the “well” of the nuclear potential). This is the only plausible choice.

   4. Option (D): 0.3 fm …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If a thorium nucleus of mass number 232 and atomic number 90 emits six alpha particles and four $\beta^-$ particles, then the ratio of the number of protons and the number of neutrons in the resulting final nucleus is (A) $39 : 65$ (B) $41 : 63$ (C) $41 : 104$ (D) $63 : 104$

   ›Reveal solutionSolution

   Each alpha decay reduces mass by 4 and charge by 2; each β⁻ decay increases charge by 1 without changing mass. Starting from thorium-232 (90 protons, 142 neutrons), after 6 α and 4 β⁻ decays we get 82 protons and 126 neutrons, giving a ratio of 41 : 63. The correct option is (B).

   ---

   Concept & Intuition

   Nuclear decay changes the composition of the nucleus in predictable ways. An alpha particle is a helium nucleus (2 protons + 2 neutrons), so emitting one reduces both the proton count and the neutron count by 2. A β⁻ particle is an electron emitted when a neutron converts into a proton (n → p + e⁻ + ν̄), so it increases the proton count by 1 and decreases the neutron count by 1, leaving the mass number unchanged.

   We track the net change in protons (Z) and neutrons (N) from the original thorium nucleus to find the final ratio Z : N.

   ---

   Step-by-step solution

   1. Initial composition of thorium-232

      Atomic number (protons) $Z_0 = 90$

      Mass number $A_0 = 232$

      Neutrons $N_0 = A_0 - Z_0 = 232 - 90 = 142$

   2. Effect of one alpha decay

      • Protons decrease by 2
      • Neutrons decrease by 2
      • Mass number decreases by 4

   3. Effect of one β⁻ decay

      • Protons increase by 1
      • Neutrons decrease by 1
      • Mass number unchanged

   4. Apply 6 alpha decays

      Change in protons: $\Delta Z_\alpha = 6 \times (-2) = -12$

      Change in neutrons: $\Delta N_\alpha = 6 \times (-2) = -12$

      New mass number: $A = 232 - 6 \times 4 = 232 - 24 = 208$

   5. Apply 4 β⁻ decays

      Change in protons: $\Delta Z_\beta = 4 \times (+1) = +4$

      Change in neutrons: $\Delta N_\beta = 4 \times (-1) = -4$

      Mass number remains 208. …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If a graph is drawn between the binding energy per nucleon and the mass number of nucleus, then the mass number of the nucleus for which the binding energy per nucleon has a maximum value is (A) 56 (B) 35 (C) 106 (D) 115

   ›Reveal solutionSolution

   The binding energy per nucleon peaks around iron-56, so the mass number with the maximum binding energy per nucleon is 56. The correct option is (A).

   The key concept here is the binding energy per nucleon curve — a graph that plots the average energy holding a nucleus together (per proton or neutron) against the mass number $A$. This curve is fundamental in nuclear physics because it explains why certain nuclei are more stable than others and why energy is released in both fission and fusion.

   Why this approach works:

   The curve rises sharply for light nuclei, peaks near $A \approx 56$ (iron-56), then slowly declines for heavier nuclei. The peak corresponds to the most tightly bound, most stable nucleus. So, to answer the question, you simply recall the well-known shape of this curve and the mass number at its maximum.

   Let’s walk through the reasoning step by step:

   1. Recall the shape of the binding energy per nucleon curve

      • For very light nuclei (e.g., hydrogen, helium), the binding energy per nucleon is low because the strong nuclear force is not fully saturated.
      • As $A$ increases, the binding energy per nucleon rises rapidly, reaching a broad maximum around $A \approx 50$–$60$.
      • Beyond that, it gradually decreases because the repulsive Coulomb force between protons grows faster than the strong force can compensate.

   2. Identify the exact peak

      • The maximum binding energy per nucleon is about 8.8 MeV and occurs for iron-56 ($A = 56$). …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If a nucleus P converts into a nucleus Q by the decay of one alpha particle and two $\beta^-$ particles, then the nuclei P and Q are (A) Isotopes (B) Isobars (C) Isotones (D) Isomers

   ›Reveal solutionSolution

   The key is to track how the mass number $A$ and atomic number $Z$ change during alpha and beta decays. An alpha particle reduces $A$ by 4 and $Z$ by 2; each $\beta^-$ increases $Z$ by 1 with no change in $A$. After one alpha and two $\beta^-$ decays, the final nucleus has the same mass number as the original but a different atomic number, so they are isobars.

   ---

   Why this approach works

   The question asks about the relationship between two nuclei, P and Q, after a specific sequence of radioactive decays. The terms in the options—isotopes, isobars, isotones, isomers—describe how nuclei compare in proton number $Z$ and neutron number $N$ (or mass number $A = Z + N$).

   • Isotopes: same $Z$, different $A$.
   • Isobars: same $A$, different $Z$.
   • Isotones: same $N$, different $Z$.
   • Isomers: same $Z$ and $A$, but different energy states (metastable).

   So we just need to compute the net change in $A$ and $Z$ from P to Q. The decays are:

   • Alpha decay: emits a helium nucleus ($^4_2\text{He}$), so $A \to A-4$, $Z \to Z-2$.
   • Beta-minus decay: emits an electron and an antineutrino; a neutron converts to a proton, so $A$ unchanged, $Z \to Z+1$.

   Let’s apply these step by step.

   ---

   Step-by-step reasoning

   1. Start with nucleus P

      Let P have mass number $A$ and atomic number $Z$. So we write it as $^A_Z\text{P}$.

   2. First decay: one alpha particle

      After emitting an alpha particle, the new nucleus (call it intermediate X) has:

   $A_X = A - 4, \quad Z_X = Z - 2.$

   So $X = ^{A-4}_{Z-2}\text{X}$.

   3. Second decay: first $\beta^-$ particle A beta-minus decay increases the atomic number by 1, mass number unchanged. So from X:

   $A_{Y} = A - 4, \quad Z_{Y} = (Z - 2) + 1 = Z - 1.$

   So $Y = ^{A-4}_{Z-1}\text{Y}$.

   4. Third decay: second $\beta^-$ particle Another beta-minus decay:

   $A_Q = A - 4, \quad Z_Q = (Z - 1) + 1 = Z.$

   So the final nucleus Q is $^{A-4}_{Z}\text{Q}$.

   5. Compare P and Q
      • P: $^A_Z\text{P}$ …

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5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If a nucleus P converts into a nucleus Q by the decay of one alpha particle and two $\beta^-$ particles, then the nuclei P and Q are (A) Isobars (B) Isomers (C) Isotones (D) Isotopes

   ›Reveal solutionSolution

   One $\alpha$ decay lowers $Z$ by 2 and two $\beta^-$ decays raise $Z$ by 2, so $Z$ is unchanged while $A$ drops by 4. Same $Z$, different $A$ means P and Q are isotopes — option (D).

   ---

   Concept

   Track the mass number $A$ and atomic number $Z$ through each decay:

   • $\alpha$ decay: $A \to A-4$, $Z \to Z-2$.
   • $\beta^-$ decay: $A$ unchanged, $Z \to Z+1$ (a neutron becomes a proton).

   Net change (one $\alpha$ + two $\beta^-$) …

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6. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.When an element $^{232}_{90}\mathrm{Th}$ decays into $^{208}_{82}\mathrm{Pb}$, the number of $\alpha$ and $\beta^-$ particles emitted respectively are (A) 6, 2 (B) 8, 2 (C) 6, 4 (D) 4, 8

   ›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 the two equations gives 6 α and 4 β⁻, so the correct option is (C).

   When a heavy nucleus like thorium-232 undergoes a series of radioactive decays to become lead-208, it does so by emitting alpha particles (helium nuclei, $^4_2\text{He}$) and beta-minus particles (electrons, $^0_{-1}e$). Each alpha emission reduces the mass number by 4 and the atomic number by 2. Each beta-minus emission converts a neutron into a proton, so the atomic number increases by 1 while the mass number stays the same.

   The problem gives you the starting and ending nuclei. You don’t need to know the exact decay chain — just the net change. That’s the beauty of this approach: you can find the number of α and β⁻ particles purely from the difference in A and Z.

   1. Find the net change in mass number (A).

      Initial: $A = 232$ (thorium).

      Final: $A = 208$ (lead).

      Decrease in A: $232 - 208 = 24$.

      Only α particles change the mass number (each by −4). So the number of α particles, say $x$, satisfies $4x = 24$, giving $x = 6$.

   2. Find the net change in atomic number (Z).

      Initial: $Z = 90$ (thorium).

      Final: $Z = 82$ (lead).

      Net decrease in Z: $90 - 82 = 8$.

      But α particles decrease Z by 2 each, while β⁻ particles increase Z by 1 each. Let $y$ be the number of β⁻ particles. The net change in Z is:

   $-2x + y = -8$

   (The left side is the total change: α reduces Z, β⁻ increases Z; the right side is the actual decrease of 8.)

   3. Substitute $x = 6$ and solve for $y$. $$$$

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7. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.If the energy released in the fission of one uranium nucleus is 200 MeV, then the number of fissions to take place per second to produce a power of 20 MW is (A) $25 \times 10^{17}$ (B) $6.25 \times 10^{17}$ (C) $12.5 \times 10^{17}$ (D) $3.125 \times 10^{17}$

   ›Reveal solutionSolution

   Power is energy per second. Convert 200 MeV to joules, then divide the required power (20 MW) by the energy per fission to get the number of fissions per second. The answer is $6.25 \times 10^{17}$.

   The core idea here is a simple unit conversion wrapped in a physics context. Power (in watts) is joules per second. Each fission releases a fixed amount of energy — 200 MeV. So to find how many fissions happen each second, you just divide the total energy needed per second (the power) by the energy each fission gives.

   The only real work is converting MeV to joules, because 1 eV is a tiny amount of energy: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

   1. Convert the fission energy to joules. $200 \text{ MeV} = 200 \times 10^6 \text{ eV} = 2 \times 10^8 \text{ eV}$. Multiply by the conversion factor:

   $E_{\text{fission}} = (2 \times 10^8) \times (1.6 \times 10^{-19}) = 3.2 \times 10^{-11} \text{ J}.$

   2. Express the required power in joules per second.

      $20 \text{ MW} = 20 \times 10^6 \text{ W} = 2 \times 10^7 \text{ J/s}$.

   3. Find the number of fissions per second.

      Let $n$ be the number of fissions per second. Then:

   $n \times (3.2 \times 10^{-11} \text{ J}) = 2 \times 10^7 \text{ J/s}.$

   So:

   $n = \frac{2 \times 10^7}{3.2 \times 10^{-11}} = \frac{2}{3.2} \times 10^{18} = 0.625 \times 10^{18}.$

   4. Write in the form given in the options. $0.625 \times 10^{18} = 6.25 \times 10^{17}$. …

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8. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.If mass of 5 neutrons is x, mass of 2 protons is y and the mass of the nucleus $_{10}X^{20}$ is z, then the binding energy of the nucleus X is (Speed of light in vacuum is c) (A) $[10y + 10x - z]c^2$ (B) $[z - 5y - 2x]c^2$ (C) $[5y + 2x - z]c^2$ (D) $[z - 10y - 10x]c^2$

   ›Reveal solutionSolution

   The nucleus $_{10}X^{20}$ has 10 protons ($=5y$) and 10 neutrons ($=2x$); the binding energy is the mass defect times $c^2$: $[5y + 2x - z]c^2$.

   Composition of $_{10}X^{20}$: atomic number $Z = 10$ (protons) and mass number $A = 20$, so neutrons $N = A - Z = 10$.

   Mass of the constituent nucleons:

   • Mass of $5$ neutrons is $x$ $\Rightarrow$ mass of $10$ neutrons $= 2x$.
   • Mass of $2$ protons is $y$ $\Rightarrow$ mass of $10$ protons $= 5y$.

   Total mass of separated nucleons $= 5y + 2x$. …

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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.A $_{92}^{238}\mathrm{U}$ nucleus decays to a $_{82}^{214}\mathrm{Pb}$ nucleus. The number of $\alpha$ and $\beta^-$ particles emitted are (A) 6 and 2 (B) 3 and 3 (C) 2 and 6 (D) 3 and 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 the two equations gives 6 α and 2 β⁻, so option (A) is correct.

   Concept & Intuition

   When a heavy nucleus decays through a series of α and β⁻ emissions, the mass number (A) changes only due to α particles (each α knocks off 4 nucleons). The atomic number (Z) changes by −2 for each α and +1 for each β⁻. So we can set up two independent equations — one for the change in A, one for the change in Z — and solve for the unknown numbers of α and β⁻. This is a classic “decay chain algebra” problem.

   Step-by-step reasoning

   1. Identify the initial and final nuclei

      Initial: $_{92}^{238}\mathrm{U}$ → $A_i = 238$, $Z_i = 92$

      Final: $_{82}^{214}\mathrm{Pb}$ → $A_f = 214$, $Z_f = 82$

   2. Let $x$ = number of α particles, $y$ = number of β⁻ particles

      Each α reduces mass number by 4 and atomic number by 2.

      Each β⁻ increases atomic number by 1 (mass number unchanged).

   3. Write the equation for mass number change

      Only α affects A:

   $A_f = A_i - 4x$

   Substitute:

   $214 = 238 - 4x \quad \Rightarrow \quad 4x = 24 \quad \Rightarrow \quad x = 6$

   4. Write the equation for atomic number change Both α and β⁻ affect Z:

   $Z_f = Z_i - 2x + y$

   Substitute $x = 6$:

   $$$$

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10. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.If the mass defect in nuclear fission is considered as 0.1% of the original mass, then the energy released when 1 kg of $_{92}^{235}\mathrm{U}$ undergoes fission is (A) $9 \times 10^{10}\ \mathrm{J}$ (B) $9 \times 10^{11}\ \mathrm{J}$ (C) $9 \times 10^{12}\ \mathrm{J}$ (D) $9 \times 10^{13}\ \mathrm{J}$

    ›Reveal solutionSolution

    The key idea is that the mass defect (0.1% of the original mass) is converted entirely into energy via $E = \Delta m c^2$. For 1 kg of uranium, the energy released is $9 \times 10^{13}\ \mathrm{J}$, so the correct option is (D).

    The problem is about nuclear fission, where a tiny fraction of mass is converted into a huge amount of energy. The famous equation $E = mc^2$ tells us that even a small mass loss yields enormous energy because $c^2$ is so large ($c = 3 \times 10^8\ \mathrm{m/s}$). Here, the mass defect is given as 0.1% of the original mass — that means for every kilogram of uranium, only 0.1% of it "disappears" and becomes energy. The rest remains as fission products.

    Let’s work through it step by step.

    1. Find the mass defect. The original mass is $m = 1\ \mathrm{kg}$. The mass defect is 0.1% of this:

    $\Delta m = \frac{0.1}{100} \times 1\ \mathrm{kg} = 0.001\ \mathrm{kg}.$

    So only one-thousandth of a kilogram is actually converted to energy.

    2. Apply Einstein’s mass-energy equivalence. The energy released is

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

    where $c = 3 \times 10^8\ \mathrm{m/s}$. Substitute:

    $E = 0.001 \times (3 \times 10^8)^2.$

    3. Calculate $c^2$.

    $c^2 = (3 \times 10^8)^2 = 9 \times 10^{16}\ \mathrm{m^2/s^2}.$

    4. Multiply to get the energy. $$$$

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11. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.200 MeV of energy is released when a $_{92}\textrm{U}^{235}$ nucleus undergoes fission. A nuclear reactor of power 192 MW uses Uranium-235 as fuel. Assume that all the energy generated in the reactor arises from the fission of Uranium-235. The amount of uranium consumed per minute by the nuclear reactor is nearly (A) 2.35 mg (B) 7.8 mg (C) 141 mg (D) 8.46 g

    ›Reveal solutionSolution

    The reactor’s power tells us the energy needed per second; dividing by the energy per fission gives the number of fissions per second, and multiplying by the mass of one U-235 nucleus gives the mass consumed per minute. The answer is 141 mg.

    The core idea is simple: every fission of a uranium-235 nucleus releases a fixed amount of energy (200 MeV). If the reactor delivers a steady power of 192 MW, that means it must be producing 192 million joules of energy every second. The only source of that energy is the fissioning of U-235 nuclei. So the number of fissions per second is just the total energy per second divided by the energy per fission. Once we know how many nuclei fission per second, we can find the mass of uranium consumed per second — and then scale it to one minute.

    The trick is to handle the units carefully. Energy per fission is given in MeV, but power is in watts (joules per second). So we need to convert MeV to joules. Also, the mass of a single U-235 nucleus is tiny, so we’ll use Avogadro’s number to relate the mass of one mole of U-235 to the number of nuclei in that mole.

    Let’s go step by step.

    1. Convert the energy per fission from MeV to joules.

       $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$.

       So $200 \text{ MeV} = 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11} \text{ J}$.

    2. Find the number of fissions per second.

       Power $P = 192 \text{ MW} = 192 \times 10^6 \text{ W} = 1.92 \times 10^8 \text{ J/s}$.

       Number of fissions per second $n = \frac{\text{energy per second}}{\text{energy per fission}} = \frac{1.92 \times 10^8}{3.2 \times 10^{-11}}$.

       $n = \frac{1.92}{3.2} \times 10^{19} = 0.6 \times 10^{19} = 6 \times 10^{18} \text{ fissions/s}$.

    3. Find the mass of one U-235 nucleus.

       One mole of U-235 has a mass of 235 g (since atomic mass is 235 u) and contains $6.02 \times 10^{23}$ atoms (Avogadro’s number).

       Mass of one nucleus $= \frac{235 \text{ g}}{6.02 \times 10^{23}} = \frac{235}{6.02} \times 10^{-23} \text{ g}$.

       Compute: $\frac{235}{6.02} \approx 39.04$, so mass per nucleus $\approx 3.904 \times 10^{-22} \text{ g}$.

    4. Find the mass consumed per second.

       Mass per second $= n \times \text{mass per nucleus} = (6 \times 10^{18}) \times (3.904 \times 10^{-22} \text{ g})$. …

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12. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
    Q.Uranium-235 is used as fuel in a nuclear reactor that takes 30 days to use up 2 kg of fuel. If each fission gives 185 MeV of usable energy, then the power output of the reactor is (A) 5.85 MW (B) 58.5 MW (C) 585 MW (D) 5850 MW

    ›Reveal solutionSolution

    The reactor’s power is found by converting the mass of uranium-235 into the number of atoms, multiplying by the energy per fission, and dividing by the time. The result is about 58.5 MW, which is option (B).

    The core idea here is that nuclear power comes from individual fission events. Each uranium-235 nucleus that splits releases a fixed amount of energy — 185 MeV in this problem. So to find the total energy output over 30 days, you first need to know how many uranium-235 atoms are in 2 kg of fuel. That number comes from Avogadro’s constant and the molar mass of U-235. Once you have the total energy in joules, dividing by the time in seconds gives power in watts.

    Let’s walk through it step by step.

    1. Find the number of uranium-235 atoms in 2 kg. The molar mass of U-235 is 235 g/mol, so 2 kg = 2000 g. Number of moles = $\frac{2000}{235}$. Using Avogadro’s number $N_A = 6.022 \times 10^{23}$ atoms/mol, the number of atoms is

    $N = \frac{2000}{235} \times 6.022 \times 10^{23} = \frac{2000 \times 6.022 \times 10^{23}}{235}.$

    Compute: $2000 \times 6.022 = 12044$, so

    $N = \frac{12044 \times 10^{23}}{235} \approx 51.25 \times 10^{23} = 5.125 \times 10^{24} \text{ atoms}.$

    2. Convert the energy per fission from MeV to joules. $1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}$, so

    $E_{\text{per fission}} = 185 \times 1.602 \times 10^{-13} \text{ J} = 296.37 \times 10^{-13} \text{ J} = 2.9637 \times 10^{-11} \text{ J}.$

    3. Calculate the total energy released. Multiply the number of fissions (same as number of atoms, since each atom fissions once) by the energy per fission:

    $E_{\text{total}} = (5.125 \times 10^{24}) \times (2.9637 \times 10^{-11}) \text{ J}.$

    Multiply the coefficients: $5.125 \times 2.9637 \approx 15.19$.

    Multiply the powers of ten: $10^{24} \times 10^{-11} = 10^{13}$.

    So $E_{\text{total}} \approx 15.19 \times 10^{13} \text{ J} = 1.519 \times 10^{14} \text{ J}$.

    4. Convert time to seconds. 30 days = $30 \times 24 \times 60 \times 60$ seconds. …

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