For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as:

Q: For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as:

where, R0 = empirical constant (approx 1.2~fm) A = mass number (number of nucleons) Let m be the average mass of a single nucleon (proton/neutron). - Total mass of the nucleus M approx A cdot m - Volume of the nucleus V = (4/3) pi R3 = (4/3) pi (R0 A1/3)3 = (4/3) pi R03 A Now, calculate the mass density : Since m and R0 are constant parameters, the mass density is constant and **independent of A**. Sees: "Nuclear mass density representation" → Highly dense, constant. Shortcut: Since volume V pro

Q27 jee_main_2026_22_january_morning Nuclear Size and Distance of Closest Approach

7.9 MeV

α

-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) \_\_\_\_ m.

[ 14π ε₀ = 9 \times 10⁹ Nm² / C² and electron charge = 1.6 \times 10⁻¹⁹ C]

A. $5.76 \times 10⁻¹⁴$ $5.76 \times 10^{-14}$
B. $1.44 \times 10⁻¹³$ $1.44 \times 10^{-13}$
C. $2.88 \times 10⁻¹⁴$ $2.88 \times 10^{-14}$
D. $1.69 \times 10⁻¹²$ $1.69 \times 10^{-12}$

Solution

Related Formula

r₀ = (1)/(4πε₀) (q₁ q₂)/(K)

Core Logic

Nuclear scattering setup diagram for Q27 - JEE Main 2026 Morning

Using mechanical energy conservation:

PEᵢ + KEᵢ = PEf + KEf

0 + 7.9 \times 10⁶ \times 1.6 \times 10⁻¹⁹ = (k(2e)(Ze))/(r) + 0

r = 9 \times 10⁹ \times 2 \times (1.6 \times 10⁻¹⁹)² \times 797.9 \times 10⁶ \times 1.6 \times 10⁻¹⁹ = 2.88 \times 10⁻¹⁴ m

For diameter:

D = 2r = 5.76 \times 10⁻¹⁴ m

Pattern Recognition

Sees: Alpha particle scattering + atomic number + kinetic energy. Shortcut: Equate initial kinetic energy to electrostatic potential energy at closest approach distance $r$ $r$, then double for diameter $D = 2r$ $D = 2r$. Check: Results align with option (1). ✓

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Class 12 Physics: Atoms and Nuclei

Q36 jee_main_2026_22_january_morning Nuclear Binding Energy

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4

α

particles is \_\_\_\_ kHz. [mass of He nucleus = 4.002 amu, 1 amu =

1.66 \times 10⁻²⁷

kg, h =

6.6 \times 10⁻³⁴

J.s and c =

3 \times 10⁸

m/s]

A. $9 \times 10¹⁹$ $9 \times 10^{19}$
B. $9 \times 10²⁰$ $9 \times 10^{20}$
C. $14.94 \times 10²⁰$ $14.94 \times 10^{20}$
D. $14.94 \times 10¹⁹$ $14.94 \times 10^{19}$

Solution

Related Formula

hν = Δ m \cdot c²

Core Logic

Mass defect calculation:

hν = (4 \times 4.002 - 15.348) \times 1.66 \times 10⁻²⁷ \times (3 \times 10⁸)²

ν = 14.94 \times 10¹⁹ kHz

Pattern Recognition

Sees: Minimum photon frequency required for nuclear breakup. Shortcut: Compute mass defect $Δ m$ $\Delta m$, convert to energy via $E=Δ m c²$ $E=\Delta m c^2$, and equate to $hν$ $h\nu$. Check: Matches option (4). ✓

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Class 12 Physics: Atoms and Nuclei

Q37 jee_main_2026_23_january_evening Isobars and Isotopes

Which of the following pair of nuclei are isobars of the element?

A. $²₁H and ³₁H$
B. $²³⁶₉₂U and ²³⁸₉₂U$
C. $¹⁹⁸₈₀Hg and ¹⁹⁷₇₉Au$
D. $³₁H and ³₂He$

Solution

Related Formula

Definition: Isobars are nuclei that possess the same mass number (A) but different atomic numbers (Z).

Core Logic

Checking the mass numbers for each pair: (1) Hydrogen-2 and Hydrogen-3 -> Isotopes (2) Uranium-236 and Uranium-238 -> Isotopes (3) Mercury-198 and Gold-197 -> Neither isobars nor isotopes (4) Hydrogen-3 ( $³₁H$ ${}^3_1\mathrm{H}$) and Helium-3 ( $³₂He$ ${}^3_2\mathrm{He}$) -> Both have mass number 3.

Step 1: Final Conclusion

Since $³₁H$ ${}^3_1\mathrm{H}$ and $³₂He$ ${}^3_2\mathrm{He}$ share the same mass number, they are isobars.

Pattern Recognition

Isobars = same top number. Isotopes = same bottom number. Isotones = same difference (neutrons).

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Class 12 Physics: Nuclei

Q46 jee_main_2026_23_january_evening Nuclear Fission

The average energy released per fission for the nucleus of

²³⁵₉₂U

is 190 MeV. When all the atoms of 47 g pure

²³⁵₉₂U

undergo fission process, the energy released is

α \times 10²³

MeV. The value of

α

is ____. (Avogadro Number =

6 \times 10²³

per mole)

Numerical Answer. Answer: 228 to 228

Solution

Related Formula

Number of Moles = Given MassMolar Mass

N = Moles \times NA

Etotal = N \times Eperfission

Core Logic

Given mass of U-235 = $47 g$ $47 \, \mathrm{g}$. Molar mass of U-235 = $235 g/mol$ $235 \, \mathrm{g/mol}$. Number of moles = $(47)/(235) = (1)/(5) moles$ .

Step 1: Calculate Total Atoms

Total number of U-235 atoms ( $N$ $N$):

N = (1)/(5) \times 6 \times 10²³ = 1.2 \times 10²³ atoms

Step 2: Calculate Total Energy

Etotal = 1.2 \times 10²³ \times 190 MeV

Etotal = 228 \times 10²³ MeV

Comparing with $α \times 10²³ MeV$ $\alpha \times 10^{23} \, \mathrm{MeV}$, we get $α = 228$ $\alpha = 228$.

Pattern Recognition

Direct stoichiometry-like energy yield. Mass / Molar Mass -> Atoms -> Total Energy.

Chapter Mix

Class 12 Physics: Nuclei

Q38 jee_main_2026_24_january_morning Binding Energy

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:

A. $Both Statement I and Statement II are true$
B. $Statement I is true but Statement II is false$
C. $Both Statement I and Statement II are false$
D. $Statement I is false but Statement II is true$

Solution

Core Logic

Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron ( $A \approx 56$ $A \approx 56$) 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.

Step 1: Final Conclusion

Statement I is false but Statement II is true. Therefore, option (4) is correct.

Pattern Recognition

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.

Chapter Mix

Class 12 Physics: Nuclei

For a nucleus of mass number A$A$ and radius R$R$, the mass density of nucleus can be represented as:

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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For a nucleus of mass number $A$ $A$ and radius $R$ $R$, the mass density of nucleus can be represented as: