Choose the correct nuclear process from the below options [p: proton, n: neutron, e- : electron, e+ : positron, v…

Q: Choose the correct nuclear process from the below options [p: proton, n: neutron, e- : electron, e+ : positron, v: neutrino, nu : antineutrino]

In basic beta^- emission processes, a neutron decays inside a nucleus to satisfy lepton numbers and conservation rules: Charge Balance: 0 arrow (+1) + (-1) + 0 = 0 (Conserved) Lepton Family Index: 0 arrow 0 + (+1) + (-1) = 0 (Conserved via antineutrino entry). This perfectly isolates option (1). Negative beta emission is always accompanied by an antineutrino, whereas positive positron transformation releases a regular neutrino molecule. Class 12 Physics: Nuclei

More Nuclei Previous-Year Questions — Page 2

Q36 jee_main_2026_24_january_evening Binding Energy and Stability

The binding energy for the following nuclear reactions are expressed in MeV.

₂He³ + ₀n¹ arrow ₂He⁴ + 20 MeV

₂He⁴ + ₀n¹ arrow ₂He⁵ - 0.9 MeV

X₃, X₄, X₅

denote the stability of

₂He³, ₂He⁴

and

₂He⁵

, respectively, then the correct order is:

A. $X₄ > X₅ > X₃$ $X_{4} > X_{5} > X_{3}$
B. $X₄ = X₅ = X₃$ $X_{4} = X_{5} = X_{3}$
C. $X₄ > X₅ < X₃$ $X_{4} > X_{5} < X_{3}$
D. $X₄ < X₅ < X₃$ $X_{4} < X_{5} < X_{3}$

Solution

Related Formula

Δ Q = BEproducts - BEreactants

Core Logic

Reaction 1 releases energy ( $+20 MeV$ $+20 \text{ MeV}$), meaning the product is more stable:

BEHe⁴ - BEHe³ = 20 MeV (1)

Reaction 2 absorbs energy ( $-0.9 MeV$ $-0.9 \text{ MeV}$), meaning the product has less total binding energy change but is relatively close:

BEHe⁵ - BEHe⁴ = -0.9 MeV (2)

Step 1: Compare Stabilities

From (1): $BEHe⁴$ $BE_{He^{4}}$ is significantly larger than $BEHe³$ $BE_{He^{3}}$. From (2): $BEHe⁵ = BEHe⁴ - 0.9 MeV$ , which implies $BEHe⁵$ $BE_{He^{5}}$ is slightly less than $BEHe⁴$ $BE_{He^{4}}$.

Thus, $BEHe⁴ > BEHe⁵ > BEHe³$ . The stability directly corresponds to the total binding energy here (and BE per nucleon leads to a similar ranking given small mass differences), so: $X₄ > X₅ > X₃$ $X_4 > X_5 > X_3$

Pattern Recognition

Positive Q-value means the products are structurally tighter (more stable). Negative Q-value means you lost stability. Simply tracking algebraic differences gives the hierarchical ranking.

Chapter Mix

Class 12 Physics: Nuclei

Q31 jee_main_2026_28_january_morning Nuclear Size and Density

An atom

⁸₃X

is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by:

A. $250%$ $250\%$
B. $150%$ $150\%$
C. $225%$ $225\%$
D. $900%$ $900\%$

Solution

Related Formula

$R = R₀ A1/3$ $R = R_0 A^{1/3}$ Surface area $S \propto R² \propto A2/3$ $S \propto R^2 \propto A^{2/3}$

Core Logic

Nucleons consist of protons and neutrons. Calculate the initial and final total nucleon number $A$ $A$ to determine the ratio of surface areas. Note: Electrons are not nucleons and do not contribute to the nuclear surface area.

Step 1: Initial State

Initial number of nucleons $Aᵢ = 8$ $A_i = 8$. Initial surface area $Sᵢ \propto (Aᵢ)2/3 = 82/3 = 4K$ (where K is a constant of proportionality).

Step 2: Final State

The atom absorbs 10 protons and 9 neutrons. Final number of nucleons $Af = 8 + 10 + 9 = 27$ $A_f = 8 + 10 + 9 = 27$. Final surface area $Sf \propto (Af)2/3 = 272/3 = 9K$ .

Step 3: Percentage Growth

Percentage increase in surface area:

% increase = (Sf - Sᵢ)/(Sᵢ) \times 100

= (9K - 4K)/(4K) \times 100

= (5)/(4) \times 100 = 125%

Step 4: Final Conclusion

The calculated percentage growth is $125%$ $125\%$, which is not in the options. Our Ans. (BONUS) NTA Ans. (3) which is $225%$ $225\%$ (This would imply $(9K)/(4K) = 2.25$ $\frac{9K}{4K} = 2.25$ or $225%$ $225\%$ as the ratio $Sf/Sᵢ$ $S_f/S_i$, not the growth).

Pattern Recognition

Radius scales as $A1/3$ $A^{1/3}$, Volume scales as $A$ $A$, Surface Area scales as $A2/3$ $A^{2/3}$. Watch out for wording like 'growth' vs 'final ratio' which exam bodies sometimes confuse.

Chapter Mix

Class 12 Physics: Nuclei

Q26 jee_main_2026_28_january_evening Nuclear Size

A nucleus has mass number

α

and radius

Rα

. Another nucleus has mass number

β

and radius

Rβ

. If

β = 8α

then

Rα/Rβ

is:

A. $2$ $2$
B. $8$ $8$
C. $1$ $1$
D. $0.5$ $0.5$

Solution

Related Formula

R = R₀A1/3

where, $R$ $R$ = radius of nucleus $A$ $A$ = mass number $R₀$ $R_0$ = empirical constant

Core Logic

For the first nucleus:

Rα = R₀α1/3

For the second nucleus:

Rβ = R₀β1/3

Step 1: Taking the Ratio

RαRβ = ((α)/(β))1/3

Given that $β = 8α$ $\beta = 8\alpha$, we substitute this value:

RαRβ = ((α)/(8α))1/3 = ((1)/(8))1/3 = (1)/(2) = 0.5

Pattern Recognition

Radius scales as the cube root of the mass number. If mass becomes 8 times, radius becomes $3\sqrt(8) = 2$ $\sqrt[3]{8} = 2$ times. Hence the ratio is $1/2$ $1/2$ or $0.5$ $0.5$.

Chapter Mix

Class 12 Physics: Nuclei

Q12 jee_main_2025_02_april_evening Nuclear Fusion and Binding Energy

Energy released when two deuterons

(₁H²)

fuse to form a helium nucleus

(₂He⁴)

is: (Given: Binding energy per nucleon of

₁H² = 1.1

MeV and binding energy per nucleon of

₂He⁴ = 7.0

MeV)

A. $8.1 MeV$ $8.1 \ \mathrm{MeV}$
B. $5.9 MeV$ $5.9 \ \mathrm{MeV}$
C. $23.6 MeV$ $23.6 \ \mathrm{MeV}$
D. $26.8 MeV$ $26.8 \ \mathrm{MeV}$

Solution

Related Formula

Fusion reaction:

₁H² + ₁H² ₂He⁴

Q-value (Energy Released) of a nuclear reaction:

Q = Total Binding Energy (Products) - Total Binding Energy (Reactants)

Core Logic

Let's compute the total binding energies:

Reactants: Two deuterons $(₁H²$ $({}_1\mathrm{H}^2$).
Number of nucleons in each deuteron = 2
Binding energy per nucleon = $1.1 MeV$ $1.1 \ \mathrm{MeV}$
Total Binding Energy of reactants:

BEreactants = 2 \times [2 \times 1.1 MeV] = 4.4 MeV

Products: One helium nucleus $(₂He⁴$ $({}_2\mathrm{He}^4$).
Number of nucleons = 4
Binding energy per nucleon = $7.0 MeV$ $7.0 \ \mathrm{MeV}$
Total Binding Energy of products:

BEproducts = 4 \times 7.0 MeV = 28.0 MeV

Step 1: Calculate energy released

The energy released ( $Q$ $Q$) in the fusion process is:

Q = BEproducts - BEreactants

Q = 28.0 MeV - 4.4 MeV = 23.6 MeV

Thus, the energy released is $23.6 MeV$ $23.6 \ \mathrm{MeV}$.

Pattern Recognition

Sees: Q-value of fusion from binding energy per nucleon. Trap: Confusing "Binding Energy per nucleon" with the total binding energy of the nucleus. Always multiply by the mass number $A$ $A$ first! Shortcut: $Q = (Afinal \times BEfinal) - (Ainitial \times BEinitial) = (4 \times 7.0) - (2 \times 2 \times 1.1) = 28 - 4.4 = 23.6 MeV$ .

Chapter Mix

Class 12 Physics: Nuclei

Choose the correct nuclear process from the below options [p: proton, n: neutron, e⁻$\mathbf{e}^{-}$ : electron, e⁺$\mathrm{e}^{+}$ : positron, v: neutrino, ν$\overline{\nu}$ : antineutrino]

Solution & Explanation

Core Logic

Step 1: Conservation Cross-Check

Pattern Recognition

Chapter Mix

Reference Study Guides

More Nuclei Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Compare Stabilities

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Initial State

Step 2: Final State

Step 3: Percentage Growth

Step 4: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Taking the Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate energy released

Pattern Recognition

Chapter Mix

More Nuclei Questions — jee_main_2025_28_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Choose the correct nuclear process from the below options [p: proton, n: neutron, $e⁻$ $\mathbf{e}^{-}$ : electron, $e⁺$ $\mathrm{e}^{+}$ : positron, v: neutrino, $ν$ $\overline{\nu}$ : antineutrino]