The chapter "Nuclei" combines quantitative techniques (mass–energy equivalence, binding energy, Q–value calculations) with conceptual understanding (stability, decay modes, tunnelling) that are repeatedly tested in CBSE board exams and competitive tests like JEE/NEET. Mastery of this chapter trains students to link nuclear masses, energetics and kinetics—skills essential for solving multi-step numerical and reasoning problems.
Beyond routine calculations, competitive problems often mix models (semi-empirical mass formula), decay-chain kinetics and graphical/data interpretation (binding-energy curves, semi-log decay plots). Practising such integrated MCQs builds the pattern-recognition and calculation accuracy needed under exam time pressure.
15
Minutes
10
Questions
1 / -0
Marking
Q1. A nucleus has measured nuclear mass . Given , and , the binding energy per nucleon of is closest to:
Q2. For the fusion reaction , given , , and with , the released energy is approximately:
Q3. Assertion (A): In nuclear decay the emitted electron is one of the orbital electrons of the parent atom.
Reason (R): The electron is created inside the nucleus during the weak interaction and is therefore not an orbital electron.
Both A and R are true and R explains A
Both A and R are true but R does not explain A
A is true but R is false
A is false but R is true
Q4. A parent nucleus P (half-life ) decays to daughter D (half-life ). Initially and . The time (in days) when the activities of parent and daughter become equal is closest to:
Q5. For three isobars with the same mass number the binding energies are given as , and . Which of the following is the most likely decay tendency of the nucleus ?
will undergo decay to
will undergo (or electron capture) to
is the most stable of the three isobars
will spontaneously emit a neutron
Q6. A nucleus splits into two fragments of mass numbers and . If the average binding energies per nucleon are for the fragment, for the fragment and for , the energy released in this fission (approximately) is:
Q7. A plot of vs time for a sample is a straight line of slope for the first few days, then at late times it becomes a straight line of slope . The best physical explanation is:
A single isotope whose decay constant slowly changed due to chemistry
A mixture of two radioactive isotopes with decay constants (short-lived) and (long-lived); the fast one dominates early and the slow one later
Detector efficiency/dead-time changed midway producing an apparent slope change
A parent–daughter in secular equilibrium producing two slopes as shown
Q8. Using a simplified semi-empirical mass formula keeping only Coulomb and asymmetry terms one finds the that maximizes binding energy for a given approximately as
For , with and , the nearest integer value of the most stable is:
Q9. Assertion (A): For -emitters the logarithm of the decay constant shows an approximately linear dependence on (Geiger–Nuttall relation).
Reason (R): Quantum tunnelling through the Coulomb barrier leads to a transmission probability (Gamow factor) , which yields the observed dependence of on .
Both A and R are true and R explains A
Both A and R are true but R does not explain A
A is true but R is false
A is false but R is true
Q10. In a lab frame a neutron (mass ) strikes a stationary nucleus of mass to induce an endothermic reaction with . Using the threshold relation , the minimum kinetic energy of the neutron required is approximately: