Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II : For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon. In the light of the above statements, choose the correct answer from the options given below:

Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron (A approx 56) and then gradually decreases for heavier nuclei. Hence, Statement I is false. Statement II: Nuclei inherently tend toward states of greater stability. Thus, nuclei with lower binding energy per nucleon undergo reactions (fission or fusion) to form nuclei with higher binding energy per nucleon. Hence, Statement II is true. Statement I is false but Statement II is true. Therefore, option (4) is correct. The binding energy curve is a classic plateau curve peaking at Iron-56. Statements claiming monotonic trends across all masses in nuclear physics are generally false. Class 12 Physics: Nuclei

Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II : For all elements, nuclei with less binding en

Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron (A approx 56) and then gradually decreases for heavier nuclei. Hence, Statement I is false. Statement II: Nuclei inherently tend toward states of greater stability. Thus, nuclei with lower binding energy per nucleon undergo reactions (fission or fusion) to form nuclei with higher binding energy per nucleon. Hence, Statement II is true. Statement I is false but Statement II is t

Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II : For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon. In the light of the above statements, choose the correct answer from the options given below:

Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron (A approx 56) and then gradually decreases for heavier nuclei. Hence, Statement I is false. Statement II: Nuclei inherently tend toward states of greater stability. Thus, nuclei with lower binding energy per nucleon undergo reactions (fission or fusion) to form nuclei with higher binding energy per nucleon. Hence, Statement II is true. Statement I is false but Statement II is true. Therefore, option (4) is correct. The binding energy curve is a classic plateau curve peaking at Iron-56. Statements claiming monotonic trends across all masses in nuclear physics are generally false. Class 12 Physics: Nuclei

Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II : For all elements, nuclei with less binding en

Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron (A approx 56) and then gradually decreases for heavier nuclei. Hence, Statement I is false. Statement II: Nuclei inherently tend toward states of greater stability. Thus, nuclei with lower binding energy per nucleon undergo reactions (fission or fusion) to form nuclei with higher binding energy per nucleon. Hence, Statement II is true. Statement I is false but Statement II is t

Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding…

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

If

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

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

Core Logic

Step 1: Final Conclusion

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

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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