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 If X3, X4, X5 denote the stability of 2He3, 2He4 and 2He5 , respectively, then the correct order is:

Reaction 1 releases energy (+20 MeV), meaning the product is more stable: Reaction 2 absorbs energy (-0.9 MeV), meaning the product has less total binding energy change but is relatively close: From (1): BEHe^4 is significantly larger than BEHe^3. From (2): BEHe^5 = BEHe^4 - 0.9 MeV, which implies BEHe^5 is slightly less than BEHe^4. Thus, BEHe^4 > BEHe^5 > BEHe^3. 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 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. Class 12 Physics: Nuclei

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 If X3, X4, X5 denote the stability of 2He3, 2He4 and 2He5 ,

Reaction 1 releases energy (+20 MeV), meaning the product is more stable: Reaction 2 absorbs energy (-0.9 MeV), meaning the product has less total binding energy change but is relatively close: From (1): BEHe^4 is significantly larger than BEHe^3. From (2): BEHe^5 = BEHe^4 - 0.9 MeV, which implies BEHe^5 is slightly less than BEHe^4. Thus, BEHe^4 > BEHe^5 > BEHe^3. The stability directly corresponds to the total binding energy here (and BE per nucleon leads to a similar ranking given small mass

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 If X3, X4, X5 denote the stability of 2He3, 2He4 and 2He5 , respectively, then the correct order is:

Reaction 1 releases energy (+20 MeV), meaning the product is more stable: Reaction 2 absorbs energy (-0.9 MeV), meaning the product has less total binding energy change but is relatively close: From (1): BEHe^4 is significantly larger than BEHe^3. From (2): BEHe^5 = BEHe^4 - 0.9 MeV, which implies BEHe^5 is slightly less than BEHe^4. Thus, BEHe^4 > BEHe^5 > BEHe^3. 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 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. Class 12 Physics: Nuclei

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 If X3, X4, X5 denote the stability of 2He3, 2He4 and 2He5 ,

Reaction 1 releases energy (+20 MeV), meaning the product is more stable: Reaction 2 absorbs energy (-0.9 MeV), meaning the product has less total binding energy change but is relatively close: From (1): BEHe^4 is significantly larger than BEHe^3. From (2): BEHe^5 = BEHe^4 - 0.9 MeV, which implies BEHe^5 is slightly less than BEHe^4. Thus, BEHe^4 > BEHe^5 > BEHe^3. The stability directly corresponds to the total binding energy here (and BE per nucleon leads to a similar ranking given small mass

The binding energy for the following nuclear reactions are expressed in MeV. 2He3 + 0n1 arrow 2He4 + 20 MeV 2He4 + 0n1…

More Nuclei Previous-Year Questions — Page 2

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

Q15 jee_main_2025_02_april_morning Bohr's Atomic Model

Considering Bohr's atomic model for hydrogen atom: (A) the energy of H atom in ground state is same as energy of

He^+

ion in its first excited state. (B) the energy of H atom in ground state is same as that for

Li⁺⁺

ion in its second excited state. (C) the energy of H atom in its ground state is same as that of

He^+

ion for its ground state. (D) the energy of

He^+

ion in its first excited state is same as that for

Li⁺⁺

in its ground state Choose the correct answer from the options given below:

A. (B), (D) only
B. (A), (B) only
C. (A), (D) only
D. (A), (C) only

Solution

Related Formula

Eₙ = -13.6 (Z²)/(n²)~eV

where $Z$ $Z$ is the atomic number and $n$ $n$ is the principal quantum number.

Core Logic

Let's check each statement by calculating energy values:

H atom ground state ( $Z = 1, n = 1$ $Z = 1, n = 1$):

EH = -13.6 (1²)/(1²) = -13.6~eV

Statement (A): $He^+$ $\mathrm{He}^+$ ion in its first excited state ( $Z = 2, n = 2$ $Z = 2, n = 2$):

EHe^+, n=2 = -13.6 (2²)/(2²) = -13.6~eV

This is equal to $EH$ $E_H$. Thus, Statement (A) is correct.

Statement (B): $Li⁺⁺$ $\mathrm{Li}^{++}$ ion in its second excited state ( $Z = 3, n = 3$ $Z = 3, n = 3$):

ELi⁺⁺, n=3 = -13.6 (3²)/(3²) = -13.6~eV

This is equal to $EH$ $E_H$. Thus, Statement (B) is correct.

Statement (C): $He^+$ $\mathrm{He}^+$ ion in its ground state ( $Z = 2, n = 1$ $Z = 2, n = 1$):

EHe^+, n=1 = -13.6 (2²)/(1²) = -54.4~eV \neq EH

Thus, Statement (C) is incorrect.

Statement (D): $He^+$ $\mathrm{He}^+$ in first excited state ( $Z=2, n=2$ $Z=2, n=2$) versus $Li⁺⁺$ $\mathrm{Li}^{++}$ in ground state ( $Z=3, n=1$ $Z=3, n=1$):

ELi⁺⁺, n=1 = -13.6 (3²)/(1²) = -122.4~eV \neq -13.6~eV

Thus, Statement (D) is incorrect.

Step 1: Final Conclusion

Statements (A) and (B) are the only correct statements.

Pattern Recognition

Bohr energy is proportional to $(Z²)/(n²)$ $\frac{Z^2}{n^2}$. If the ratio of $(Z)/(n)$ $\frac{Z}{n}$ is the same for two systems, their quantum state energy levels will be identical.

Chapter Mix

Class 12 Physics: Atoms

Solution & Explanation

Related Formula

Core Logic

Step 1: Compare Stabilities

Pattern Recognition

Chapter Mix

Reference Study Guides

More Nuclei Previous-Year Questions — Page 2

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

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More Nuclei Questions — jee_main_2026_24_january_evening

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