Group
Group
Nuclei Formulas for JEE 2027
Nuclei Formulas for JEE 2027 cover nuclear radius, mass defect, binding energy, radioactive decay, half-life, mean life, and energy released in nuclear reactions. Understanding these concepts helps students solve JEE Nuclei Questions involving nuclear properties, decay processes, fission, and fusion. Accurate unit conversions and a clear distinction between atomic and nuclear masses are essential for solving these problems correctly.
Use this formula sheet alongside your JEE Nuclei Notes to review key relations, their conditions, and worked examples. Practise questions independently, then check the tables to identify gaps in your understanding. Attempt a Free JEE Advanced mock test to assess your preparation and revisit mistakes involving units, decay calculations, and energy signs.
Notation
| Symbol | Stands for | Units |
|---|---|---|
| $$A$$ | Mass number (total nucleons) | Dimensionless |
| $$Z$$ | Atomic number (protons) | Dimensionless |
| $$N$$ | Number of neutrons ($$A-Z$$) | Dimensionless |
| $$R$$ | Radius of a nucleus | m |
| $$r_0$$ | Empirical constant in radius relation | m |
| $$\rho$$ | Nuclear density | kg m−3 |
| $$m_p, m_n$$ | Rest mass of proton, neutron | kg or u |
| $$m_N$$ | Rest mass of neutral atom | kg or u |
| $$\Delta m$$ | Mass defect | kg or u |
| $$E_b$$ | Total binding energy | J or MeV |
| $$B.E./A$$ | Binding energy per nucleon | MeV nucleon−1 |
| $$c$$ | Speed of light $$3.00\times10^{8}$$ | m s−1 |
| $$\lambda$$ | Decay constant | s−1 |
| $$T_{1/2}$$ | Half-life | s |
| $$\tau$$ | Mean life | s |
| $$N_0$$ | Initial number of nuclei | Dimensionless |
| $$N(t)$$ | Nuclei remaining at time $$t$$ | Dimensionless |
| $$A(t)$$ | Activity at time $$t$$ | s−1 (Bq) |
| $$A_0$$ | Initial activity | Bq |
| $$Q$$ | Q-value of a nuclear reaction | J or MeV |
| $$u$$ | Unified atomic mass unit ($$1.6605\times10^{-27}$$ kg) | kg |
| $$1\ \text{u}c^{2}$$ | Energy equivalent $$931.5$$ MeV | MeV |
Nuclear Radius and Density Formulas
This block fixes the geometry: how big a nucleus is and how tightly the nucleons are packed. Most JEE problems start here to give you a handle on volume or density before moving to energy. It builds straight from the previous idea of atomic structure you learnt in Modern Physics and feeds the mass-defect calculations in the next section.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$R=r_0A^{1/3}$$ | Nuclear radius | $$r_0\approx1.2\times10^{-15}\ \text{m}$$ |
| $$V=\dfrac{4}{3}\pi R^{3}$$ | Nuclear volume | Sphere approximation |
| $$\rho=\dfrac{m_N}{V}$$ | Nuclear density | Matter in nucleus assumed uniform |
| $$\rho\approx2.3\times10^{17}\ \text{kg m}^{-3}$$ | Numerical density value | Any nucleus (variation < 2%) |
| $$r_p\approx0.8\times10^{-15}\ \text{m}$$ | Charge radius of proton | Special data point |
Worked Example
Find the radius of $${}^{208}\text{Pb}$$ nucleus and its density in kg m−3. Take $$r_0=1.2\times10^{-15}\ \text{m}$$ and $$m({}^{208}\text{Pb})=208\ \text{u}$$.
- Radius: $$R=r_0A^{1/3}=1.2\times10^{-15}\times208^{1/3}$$.
- Compute $$208^{1/3}\approx5.93$$. So $$R=1.2\times10^{-15}\times5.93=7.12\times10^{-15}\ \text{m}$$.
- Volume $$V=\dfrac{4}{3}\pi R^{3}=1.33\pi(7.12\times10^{-15})^{3}=1.33\pi(3.61\times10^{-43})$$.
- $$V\approx1.51\times10^{-42}\ \text{m}^{3}$$.
- Mass $$m_N=208u=208\times1.6605\times10^{-27}=3.45\times10^{-25}\ \text{kg}$$.
- Density $$\rho=m_N/V=3.45\times10^{-25}/1.51\times10^{-42}=2.29\times10^{17}\ \text{kg m}^{-3}$$.
Answer: $$R=7.1\times10^{-15}\ \text{m},\ \rho=2.3\times10^{17}\ \text{kg m}^{-3}$$
Shortcuts and Special Cases
- Density is constant for all nuclei – you can quote $$2.3\times10^{17}$$ without calculation unless asked for derivation.
- $$R(\text{Fe})\approx4.6\times10^{-15}\ \text{m}$$ – handy for ratio questions.
- If the nucleus doubles its mass number, radius scales by factor $$2^{1/3}=1.26$$.
Watch Out
- Using $$A^{2/3}$$ instead of $$A^{1/3}$$ – common mix-up with surface area relation.
- For density, forget to convert u to kg: 931 MeV/c² is not kg.
- r₀ value varies in books; stick to $$1.2$$ fm for JEE unless specific value is supplied.
Mass Defect and Packing Fraction Formulas
Now you quantify how mass turns into energy. The examiner usually lets you calculate binding energy through mass defect, but may wrap it into a ratio called packing fraction. That prepares the ground for stability discussions in the next block.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\Delta m=Z m_p+N m_n-m_N$$ | Mass defect of a neutral atom | Electron masses neglected if not given |
| $$E_b=\Delta m c^{2}$$ | Total binding energy | Δm in kg → J; in u → MeV via 931.5 |
| $$\dfrac{E_b}{A}$$ | Binding energy per nucleon | Units MeV/nucleon |
| $$f=\dfrac{(m_N-A)u}{A}$$ | Packing fraction $$f$$ | Approx formula: small numbers (×10−3) |
| $$1\ \text{u}c^{2}=931.5\ \text{MeV}$$ | Energy‐mass conversion constant | Always |
Worked Example
For $${}^{56}\text{Fe}$$, masses: $$m_p=1.00728\ \text{u},\ m_n=1.00866\ \text{u},\ m({}^{56}\text{Fe})=55.93494\ \text{u}$$. Find (a) mass defect, (b) binding energy per nucleon.
- $$Z=26,\ N=56-26=30$$.
- Combined nucleon mass $$=26(1.00728)+30(1.00866)=26.1893+30.2598=56.4491\ \text{u}$$.
- Mass defect $$\Delta m=56.4491-55.93494=0.51416\ \text{u}$$.
- Total binding energy $$E_b=0.51416\times931.5=479.9\ \text{MeV}$$.
- Per nucleon $$E_b/A=479.9/56=8.57\ \text{MeV}$$.
Answer: $$\Delta m=0.514\ \text{u},\ B.E./A=8.6\ \text{MeV}$$
Shortcuts and Special Cases
- Iron peak: $$B.E./A\approx8.8$$ MeV for A ≈ 56 – stability benchmark.
- Packing fraction negative means mass defect positive (stable).
- For quick checks, $$1\ \text{u}\approx931$$ MeV; difference of 0.001 u equals 0.93 MeV.
Watch Out
- Forget electrons: subtract 0.00055 u per electron only if high precision demanded; JEE rarely does.
- Plugging Δm in u into $$E=mc^{2}$$ directly without 931: gives absurdly small energy.
- Packing fraction sign convention: many students write |f|; examiner wants the sign.
Binding Energy and Stability Formulas
Once you have $$E_b$$, you decide if a nucleus will split or join. Questions test comparison – which isotope releases energy on fusion/fission – and revolve around binding energy per nucleon curves.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\Delta E=E_f-E_i$$ | Energy change in a transformation | Conservation of nucleon number |
| If $$B.E./A$$ increases ⇒ energy released | Stability criterion | Qualitative rule |
| Maximum $$B.E./A$$ at $$A\approx56$$ | Iron peak | Global curve |
| Fusion favourable for $$A\lt30$$ | Low-mass nuclei fuse | Assuming product nearer 56 |
| Fission favourable for $$A\gt200$$ | Heavy nuclei split |
Worked Example
Helium-4 nuclei combine to form carbon-12. How much energy is released per fusion reaction? Masses: $$m({}^{4}\text{He})=4.00260\ \text{u},\ m({}^{12}\text{C})=12.00000\ \text{u}$$.
- Initial mass $$=3\times4.00260=12.00780\ \text{u}$$.
- Final mass $$=12.00000\ \text{u}$$.
- Mass defect $$\Delta m=12.00780-12.00000=0.00780\ \text{u}$$.
- Energy released $$Q=\Delta m c^{2}=0.00780\times931.5=7.27\ \text{MeV}$$.
Answer: 7.3 MeV released per reaction
Shortcuts and Special Cases
- For alpha fusion to carbon, remember ~7 MeV – pops up in assertion-reason questions.
- A 1 u mass defect roughly equals energy of $$1.5\times10^{-10}$$ J.
- If energy per nucleon rises, the process is exothermic; fall means endothermic – saves calculation.
Watch Out
- Comparing total binding energies instead of per nucleon when asked about stability.
- Neglecting number of particles: 3 He-4 to 1 C-12, not 1 to 1.
- Using wrong sign for energy: released is positive Q.
Nuclear Force Characteristics Formulas
Though qualitative, JEE slips one MCQ on range or saturation. Few formulas, but you still need the scaling laws that justify constant density and binding energy trends.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Force range $$\approx1-2\ \text{fm}$$ | Effective range of strong nuclear force | Between nucleons |
| Potential $$V(r)\sim -V_0 e^{-r/r_0}$$ | Yukawa-type potential (qualitative) | $$r\lt3\ \text{fm}$$ |
| Saturation: binding energy $$\propto A$$ | Because each nucleon interacts with fixed neighbours | For bulk nuclei |
Worked Example
Explain in one calculation why density is constant using saturation concept.
- Binding energy $$E_b\propto A$$ due to saturation.
- Volume $$V\propto A$$ (from constant density assumption).
- Equate proportionalities: $$E_b/V\approx\text{constant}$$ → energy density same, supporting constant ρ.
Answer: Saturation of nuclear forces makes both binding energy and volume scale linearly with A, so density stays constant.
Shortcuts and Special Cases
- Range 1 fm – memorise; option with 10 fm is wrong nine times out of ten.
- Saturation explains $$R\propto A^{1/3}$$ quickly in one mark True/False questions.
Watch Out
- Confusing range of nuclear force (fm) with range of weak force (10−3 fm).
- Thinking nuclear force is inverse-square – it is not.
Radioactive Decay Law Formulas
This is where JEE loves to hide logarithms. Almost every year a two-step decay law question is asked. You will chain this law with half-life and activity formulas next.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$N(t)=N_0 e^{-\lambda t}$$ | Nuclei remaining after time t | Single decay constant |
| $$A(t)=\lambda N(t)$$ | Activity at time t | Instantaneous rate |
| $$\ln\dfrac{N_0}{N}= \lambda t$$ | Linear log form | |
| $$dN/dt=-\lambda N$$ | Differential equation of decay |
Worked Example
A radioactive sample shows 8000 counts per minute now and 2000 counts after 1 hour. Find $$\lambda$$.
- Activity proportional to N, so $$N_0/N=8000/2000=4$$.
- Use $$\ln4=\lambda t$$ where $$t=3600\ \text{s}$$.
- $$\ln4=1.386$$.
- $$\lambda=1.386/3600=3.85\times10^{-4}\ \text{s}^{-1}$$.
Answer: $$\lambda=3.9\times10^{-4}\ \text{s}^{-1}$$
Shortcuts and Special Cases
- Every factor-of-2 drop means one half-life; factor-of-4 is two half-lives – skips logs.
- For small $$\lambda t\lt0.1$$, $$e^{-\lambda t}\approx1-\lambda t$$ – good for screening questions.
Watch Out
- Using base-10 log with natural log constant – stick to ln.
- Counts per minute vs per second mismatch in $$\lambda$$ units.
Half-Life, Mean Life & Activity Relations Formulas
These relations convert decay constant into tangible time scales. Examiners love to mix half-life data from tables with activity calculations.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$T_{1/2}=\dfrac{\ln2}{\lambda}=0.693/\lambda$$ | Half-life | Single decay constant |
| $$\tau=1/\lambda$$ | Mean (average) life | |
| $$\tau=1.44\,T_{1/2}$$ | Relation between mean and half life | |
| $$A(t)=A_0 e^{-\lambda t}$$ | Activity decay | |
| $$A_0=\lambda N_0$$ | Initial activity |
Worked Example
An isotope has half-life 10 days and an initial activity of 5 MBq. What will be its activity after 25 days?
- Decay constant $$\lambda=0.693/10=0.0693\ \text{day}^{-1}$$.
- Activity after t days: $$A=A_0 e^{-\lambda t}=5e^{-0.0693\times25}$$.
- Compute exponent: $$0.0693\times25=1.7325$$.
- $$e^{-1.7325}=0.177$$.
- Activity $$A=5\times0.177=0.885\ \text{MBq}$$.
Answer: $$0.89\ \text{MBq}$$
Shortcuts and Special Cases
- 25 days ≈ 2.5 half-lives → activity drops by $$2^{2.5}=5.66$$. 5/5.66 ≈ 0.88 MBq – sanity check without calc.
- To convert days to seconds, multiply by 86,400 quickly using 8.64×104.
Watch Out
- Using $$A_0/A$$ instead of A/A0 in exponent sign.
- Confusing mean life with half life – they differ by 44%.
Successive Disintegration and Radioactive Equilibrium Formulas
Chain decay shows up in paragraph questions of JEE Advanced. You need mother–daughter equations and equilibrium conditions. This links directly with decay constant concepts just learnt.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$N_2(t)=\dfrac{\lambda_1 N_0}{\lambda_2-\lambda_1}(e^{-\lambda_1 t}-e^{-\lambda_2 t})$$ | Daughter nuclei at time t | $$\lambda_1\neq\lambda_2$$ |
| Secular equilibrium: $$\lambda_1 N_1=\lambda_2 N_2$$ | Condition when parent half-life ≫ daughter | |
| Transient equilibrium: $$A_2/A_1$$ tends to constant | Parent half-life a few times daughter |
Worked Example
Parent isotope $$P$$ (half-life 30 days) decays to daughter $$D$$ (half-life 3 days). After long time, what is the ratio $$A_D/A_P$$?
- Secular equilibrium not valid (half-lives differ by factor 10, still large). Use activity ratio limit for long time: $$A_D/A_P=\lambda_P/\lambda_D$$.
- Compute decay constants: $$\lambda_P=0.693/30=0.0231\ \text{day}^{-1}$$, $$\lambda_D=0.693/3=0.231\ \text{day}^{-1}$$.
- Ratio $$=0.0231/0.231=0.10$$.
Answer: Activity of daughter is 0.10 of parent in steady state.
Shortcuts and Special Cases
- If parent half-life is >20× daughter, use secular equilibrium $$A_D=A_P$$ – saves long derivation.
- Chain of more than two rarely asked; if it appears treat earlier links as secular.
Watch Out
- Swapping λ’s in ratio – keep larger λ in denominator for activity ratio.
- For identical λ, formula blows up – handle separately using limit $$N_2=\lambda N_0 te^{-\lambda t}$$.
Q-Value of Nuclear Reactions Formulas
Energy bookkeeping for any nuclear reaction. Examiner may couple it with conservation of momentum to ask for kinetic energies. This section relies on mass defect formulas already seen.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$Q=(\sum m_{\text{initial}}-\sum m_{\text{final}})c^{2}$$ | Net energy released (+) or absorbed (−) | Rest masses in u or kg |
| Exoergic if $$Q\gt0$$ | Reaction releases energy | |
| Endoergic if $$Q\lt0$$ | Requires threshold energy | |
| Threshold energy $$E_{th}=-Q\left(1+\dfrac{m_{\text{projectile}}}{m_{\text{target}}}\right)$$ | Minimum KE for endoergic reaction in lab frame | Non-relativistic limit |
Worked Example
Calculate the Q-value of $$\alpha + {}^{14}\text{N} \rightarrow {}^{17}\text{O}+p$$. Atomic masses: $$\alpha=4.00260\ \text{u},\ ^{14}\text{N}=14.00307\ \text{u},\ ^{17}\text{O}=16.99913\ \text{u},\ p=1.00728\ \text{u}$$.
- Total initial mass $$=4.00260+14.00307=18.00567\ \text{u}$$.
- Total final mass $$=16.99913+1.00728=18.00641\ \text{u}$$.
- Δm $$=18.00567-18.00641=-0.00074\ \text{u}$$.
- $$Q=-0.00074\times931.5=-0.69\ \text{MeV}$$ (endoergic).
Answer: $$Q=-0.69\ \text{MeV}$$ (reaction needs 0.69 MeV input)
Shortcuts and Special Cases
- If masses differ by 0.001 u, energy ≈ 0.93 MeV – quick mental check.
- For fission of U-235 into Ba-141 + Kr-92 + 3n, Q ≈ 200 MeV – number worth knowing.
Watch Out
- Incorrect sign of Δm – always initial minus final.
- Using atomic instead of nuclear masses when electrons imbalance between sides – they cancel if equal Z both sides; else adjust.
Nuclear Fission & Fusion Energy Calculations Formulas
The heavy end and the light end of binding energy curve. JEE sometimes asks how many fissions light a bulb or power a reactor. Good place to fetch easy marks if you carry the constants.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Fission energy per event $$\approx200\ \text{MeV}$$ for $$^{235}\text{U}$$ | Rule-of-thumb value | Thermal neutron fission |
| Energy released $$E=n Q$$ | Total energy for n events | Independent events |
| Number of fissions $$n=\dfrac{P t}{Q}$$ | Given power P over time t | Continuous operation |
| Fusion of $$D+T$$ releases $$17.6\ \text{MeV}$$ | Standard fusion reaction | |
| Mass-energy conversion efficiency $$\eta=E/(m c^{2})$$ | Fraction of mass converted to energy |
Worked Example
How much $$^{235}\text{U}$$ is consumed in a 1 GW power plant operating at 30 % thermal efficiency for 1 day? Assume 200 MeV per fission.
- Electrical energy output $$E_{\text{out}}=1\ \text{GW}\times86400\ \text{s}=8.64\times10^{13}\ \text{J}$$.
- Thermal energy required $$E_{\text{th}}=E_{\text{out}}/0.30=2.88\times10^{14}\ \text{J}$$.
- Energy per fission $$Q=200\ \text{MeV}=200\times1.602\times10^{-13}=3.20\times10^{-11}\ \text{J}$$.
- Number of fissions $$n=E_{\text{th}}/Q=2.88\times10^{14}/3.20\times10^{-11}=9.0\times10^{24}$$.
- Mass of one $$^{235}\text{U}$$ nucleus $$=235u=235\times1.6605\times10^{-27}=3.90\times10^{-25}\ \text{kg}$$.
- Total mass consumed $$m=n\times3.90\times10^{-25}=3.5\times10^{0}\ \text{kg}$$.
Answer: About 3.5 kg of U-235 per day
Shortcuts and Special Cases
- Rule: 1 MW day ≈ 1 g of U-235 burnt – easy scaling.
- Fusion: 1 g of D-T mixture releases ≈ 340 GJ.
Watch Out
- Power vs energy confusion: multiply by time in seconds.
- Efficiency applied the wrong way (thermal vs electrical).
Commonly Confused Nuclei Formulas for JEE 2027
| Confusing Pair | Where They Belong | How to Choose Correctly |
|---|---|---|
| $$R=r_0A^{1/3}$$ vs $$T_{1/2}=0.693/\lambda$$ | Nuclear size / Radioactivity | R involves mass number, T1/2 involves decay constant; check dimension. |
| $$E_b=\Delta m c^{2}$$ vs $$Q=(m_i-m_f)c^{2}$$ | Single nucleus / Reaction | Binding energy uses constituents of SAME nucleus; Q uses reactants and products set. |
| $$\tau=1/\lambda$$ vs $$\tau=1.44 T_{1/2}$$ | Mean life | If λ given, use first; if half-life given, use second. |
| $$N(t)=N_0e^{-\lambda t}$$ vs $$N_2(t)=\dfrac{\lambda_1N_0}{\lambda_2-\lambda_1}(e^{-\lambda_1 t}-e^{-\lambda_2 t})$$ | Single decay / Chain decay | Look for daughter nucleus mention; if yes, second formula. |
Common Mistakes to Avoid in JEE Nuclei Formulas
- Forgetting to convert unified mass unit to kg or MeV consistently.
- Plugging base-10 logarithm into decay laws that need natural log.
- Using binding energy per nucleon directly to calculate Q without multiplying by A.
- Dropping electron masses incorrectly when Z changes across reaction.
- Assuming nuclear force is inverse-square; it is short-range and saturating.
- Mistaking half-life for mean life, losing 44 % in answer.
- Neglecting efficiency factors in fission power problems.
- Swapping initial and final masses, flipping sign of Q-value.
Quick Revision of Nuclei Formulas for JEE 2027
- $$R=1.2\times10^{-15}A^{1/3}\ \text{m}$$ – radius scaling.
- $$E_b=\Delta m c^{2}$$ with $$1\ \text{u}=931.5\ \text{MeV}$$ – convert mass to energy.
- $$T_{1/2}=0.693/\lambda$$ – half-life-decay constant link.
- $$N(t)=N_0e^{-\lambda t}$$ – decay law.
- $$\tau=1.44T_{1/2}$$ – mean life shortcut.
- Q-value: $$Q=(m_i-m_f)c^{2}$$ – positive means energy out.
- Fission of U-235 releases ≈200 MeV – remember the number.
- Density of nucleus ≈ $$2.3\times10^{17}\ \text{kg m}^{-3}$$ – universal constant.
- In chain equilibrium, $$A_d/A_p=\lambda_p/\lambda_d$$ – ratio formula.
- Fusion energy D+T = 17.6 MeV – ready number.
Nuclei Formulas for JEE 2027: Conclusion
Nuclei Formulas for JEE 2027 bring together nuclear size, mass defect, binding energy, radioactive decay, and nuclear reaction calculations. Effective revision means understanding what each quantity represents and when each relation applies. Keep this chapter in your JEE Physics Formula Sheet, with clear notes on units, atomic and nuclear masses, and the distinction between half-life and mean life.
Pair this sheet with your JEE Nuclei Notes and solve a varied set of JEE Nuclei Questions before attempting a Free JEE Advanced mock test. Review every incorrect answer and record the concept or calculation that caused the mistake. As you revise JEE Physics Important Chapters, revisit nuclei regularly to strengthen your understanding and improve accuracy in Modern Physics questions.
.webp)
