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The Vault
RowQ
The Vault
CBSE Class 12 Physics · 11 questions · 26 marks
This chapter looks inside the nucleus itself, at the protons and neutrons packed together by a force far stronger than electric repulsion, and at the mass-energy bookkeeping that explains why splitting or joining nuclei releases enormous amounts of energy. It also covers the statistics of radioactive decay, where individual nuclei disintegrate at random but a large sample follows a precise exponential law.
The radius of a nucleus is related to its mass number A by R = R₀A^(1/3). This relation implies that nuclear density is:
Answer
Nearly the same (constant) for all nuclei — since volume scales as R³ = R₀³A, and mass scales as A (roughly, in nucleon mass units), density = mass/volume is independent of A, meaning nuclear matter is packed at essentially the same density throughout the periodic table.
In beta-minus decay, a nucleus emits an electron. As a result, the daughter nucleus has:
Answer
Z increased by 1, A unchanged — beta-minus decay occurs when a neutron inside the nucleus converts into a proton, an electron, and an antineutrino; the proton count (Z) rises by one while the total nucleon count (A) stays the same, since a neutron simply became a proton.
A radioactive sample has a half-life of 20 days. The fraction of the original sample remaining undecayed after 60 days is:
Answer
1/8 — 60 days is exactly 3 half-lives (60/20 = 3), and the surviving fraction after n half-lives is (1/2)ⁿ = (1/2)³ = 1/8 of the original sample.
The binding energy per nucleon curve shows a peak around mass number A ≈ 56. This is the physical reason why:
Answer
Both fission of heavy nuclei and fusion of light nuclei release energy — nuclei on either side of the peak (A ≈ 56, near iron) have lower binding energy per nucleon than nuclei closer to the peak, so heavy nuclei splitting apart or light nuclei joining together both move products toward the more tightly bound middle region, releasing the difference in binding energy.
Assertion (A): The mass of a nucleus is always less than the sum of the masses of its constituent free protons and neutrons. Reason (R): Some mass is converted into binding energy that holds the nucleons together, according to E = Δmc².
Answer
Both A and R are true and R is the correct explanation of A — when nucleons bind together into a nucleus, energy is released (the binding energy), and by mass-energy equivalence this released energy corresponds exactly to the mass defect, the amount by which the nucleus's mass falls short of the sum of its free constituent masses.
Define half-life and mean life of a radioactive sample, and state the relation between them.
Answer
Half-life (T½) is the time in which exactly half of the radioactive nuclei in a given sample decay. Mean life (τ) is the average lifetime of a nucleus before it decays, equal to the reciprocal of the decay constant, τ = 1/λ. The relation between them is T½ = 0.693τ (equivalently τ = T½/0.693 = 1.44 T½), so the mean life is always somewhat longer than the half-life.
A radioactive isotope has a decay constant λ = 0.0231 per day. Calculate its half-life, and find what fraction of a sample remains undecayed after 90 days.
Answer
Half-life: T½ = 0.693/λ = 0.693/0.0231 = 30 days. 90 days corresponds to 90/30 = 3 half-lives. Fraction remaining = (1/2)³ = 1/8. So after 90 days, one-eighth of the original radioactive sample remains undecayed, and the rest has transformed into the daughter product.
Explain why nuclear fusion, though it releases more energy per unit mass than fission, has not yet been harnessed for controlled power generation on Earth as fission has.
Answer
Fusion requires two positively charged light nuclei to be pushed close enough together (within nuclear force range, about 1 fm) to fuse, but they strongly repel each other electrostatically at larger distances. Overcoming this Coulomb barrier needs extremely high temperatures (tens of millions of kelvin, as in the Sun's core) and densities to be maintained in a stable, confined state long enough for net energy gain, a combination of extreme temperature and confinement that has been technologically very difficult to sustain and control on Earth, unlike fission which proceeds readily once a critical mass of heavy nuclei is assembled.
Derive the law of radioactive decay N(t) = N₀e^(-λt) starting from the basic assumption that the decay rate is proportional to the number of undecayed nuclei present. Use it to find the activity of a sample after 2 half-lives if its initial activity was 800 disintegrations per second.
Answer
The basic assumption: at any instant, the number of nuclei decaying per unit time is proportional to the number of undecayed nuclei present, dN/dt = -λN, where λ is the decay constant and the negative sign shows N decreasing. Separating variables: dN/N = -λ dt. Integrating from t = 0 (N = N₀) to time t (N = N(t)): ln(N/N₀) = -λt. Exponentiating both sides: N(t) = N₀e^(-λt), the exponential decay law. Activity A = |dN/dt| = λN, so activity also follows A(t) = A₀e^(-λt), and after each half-life the activity halves just like N does. Given A₀ = 800 dis/s, after 2 half-lives: A = A₀ × (1/2)² = 800 × 1/4 = 200 disintegrations per second.
(a) Explain what is meant by a nuclear chain reaction and the role of a moderator and control rods in a nuclear reactor. (b) In a fission reaction, ²³⁵U absorbs a neutron and splits into two fragments with total binding energy per nucleon higher than that of ²³⁵U by about 0.9 MeV per nucleon on average. Estimate the total energy released per fission event.
Answer
(a) A chain reaction occurs when each fission event releases neutrons (typically 2-3 per fission of ²³⁵U) that go on to trigger further fission events in neighbouring nuclei, so the number of reactions multiplies rapidly if left unchecked. A moderator (commonly water or graphite) slows down the fast neutrons produced in fission to thermal speeds, since slow neutrons are far more likely to be captured and cause further fission in ²³⁵U. Control rods (made of neutron-absorbing material like boron or cadmium) are inserted or withdrawn to absorb excess neutrons, keeping the chain reaction steady (critical) rather than runaway (supercritical), allowing controlled, sustained power generation. (b) ²³⁵U has mass number A = 235. An increase in binding energy per nucleon of 0.9 MeV, spread over all 235 nucleons that end up in the fragments, gives total energy released ≈ 235 × 0.9 MeV = 211.5 MeV per fission event. This is consistent with the well-known figure of roughly 200 MeV released per fission of ²³⁵U, most of which appears as kinetic energy of the fission fragments.
Read the passage and answer the questions that follow: A sample of a radioactive isotope used in a hospital scanner has a half-life of 6 hours. At 8:00 AM, a fresh sample is prepared with an activity of 640 units, enough for the day's scheduled scans. (a) Write the decay constant λ of this isotope in terms of its half-life. (b) Calculate the activity of the sample at 8:00 PM the same day. (c) Calculate how many half-lives have elapsed by 2:00 AM the next day, and the corresponding activity. (d) Explain, in terms of the decay law, why the sample's activity never mathematically reaches exactly zero.
Answer
(a) λ = 0.693/T½ = 0.693/6 = 0.1155 per hour. (b) From 8:00 AM to 8:00 PM is 12 hours = 12/6 = 2 half-lives. Activity = 640 × (1/2)² = 640/4 = 160 units. (c) From 8:00 AM to 2:00 AM next day is 18 hours = 18/6 = 3 half-lives. Activity = 640 × (1/2)³ = 640/8 = 80 units. (d) The decay law N(t) = N₀e^(-λt) is a continuous exponential function that approaches zero only as t approaches infinity; for any finite time t, e^(-λt) is a positive number greater than zero, so mathematically some (extremely small, eventually less than one nucleus in a real sample) activity always remains in the idealised continuous model.
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