Atoms and Nuclei Previous Year Questions — JEE Main Physics
23 past-year Atoms and Nuclei questions from JEE Main (Physics) — page 2.
Q1 (2025)
A radioactive material P first decays into Q and then Q decays to non-radioactive material R. Which of the following figure represents time dependent mass of P, Q and R?
- {{IMG1}}
- {{IMG2}}
- {{IMG3}}
- {{IMG4}}
### Related Formula
$$N = N_0 e^{-\lambda t}$$
where $\lambda$ is the decay constant.
### Core Logic
Initially, only material P is present, so its mass decreases exponentially from a maximum value to zero.
Material Q is formed from P and then decays into R, so its mass initially increases from zero, reaches a maximum, and then decreases to zero.
Material R is stable and accumulated over time, so its mass increases continuously from zero and levels off at a maximum value equal to the initial mass of P.
### Step 1: Graphical Identification
Looking at the options, option (2) correctly depicts the exponential decay of P, the transient rise and fall of Q, and the continuous growth of R to a stable value. {{SOL_IMG1}}
### Pattern Recognition
For sequential decay $P \rightarrow Q \rightarrow R$, parent $P$ always starts at max and drops to 0. Intermediate $Q$ starts at 0, peaks, and returns to 0. Final stable product $R$ starts at 0 and grows asymptotically to max value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q20 (2025)
Considering the Bohr model of hydrogen like atoms, the ratio of the radius $5^{\mathrm{th}}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$ is
- $\frac{3}{2}$
- $\frac{4}{9}$
- $\frac{9}{4}$
- $\frac{2}{3}$
### Related Formula
Bohr orbit radius expression:
$$r_n \propto \frac{n^2}{Z}$$
For a constant orbit index ($n = 5$):
$$r_5 \propto \frac{1}{Z}$$
where $Z$ is the atomic atomic number identifier.
### Core Logic
Identify the atomic values:
* For $\mathrm{Li}^{2+} implies Z_{\mathrm{Li}} = 3$
* For $\mathrm{He}^{+} implies Z_{\mathrm{He}} = 2$
### Step 1: Compute Ratio
Set up the inverse scaling ratio format:
$$\frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{Z_{\mathrm{He}}}{Z_{\mathrm{Li}}} = \frac{2}{3}$$
### Pattern Recognition
Since orbit numbers are identical, the radius is purely inversely proportional to the atomic nuclear charge $Z$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q2 (2025)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) The density of the copper $(\binom{64}{29}\mathrm{Cu})$ nucleus is greater than that of the carbon $(\binom{12}{6}\mathrm{C})$ nucleus. [cite: 20]
Reason (R): The nucleus of mass number A has a radius proportional to $\mathrm{A}^{1/3}$. [cite: 21]
In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 22]
- (A) is correct but (R) is not correct [cite: 23]
- (A) is not correct but (R) is correct [cite: 24]
- Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 25]
- Both (A) and (R) are correct but (R) is not the correct explanation of (A) [cite: 26]
### Related Formula
$R = R_0 A^{1/3}$ [cite: 667]
$$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m_n A}{\frac{4}{3}\pi R^3}$$ [cite: 664]
### Core Logic
Substituting the expression for radius $R$ into the density equation: [cite: 664]
$$\rho = \frac{m_n A}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{m_n A}{\frac{4}{3}\pi R_0^3 A} = \frac{m_n}{\frac{4}{3}\pi R_0^3}$$ [cite: 664]
As observed, the mass number $A$ cancels out perfectly, implying that the density of all nuclei is roughly identical and constant regardless of their mass numbers[cite: 664, 666]. Thus, the nuclear density of copper is equal to that of carbon, meaning Assertion (A) is incorrect[cite: 20, 663]. Reason (R) is correct since $R \propto A^{1/3}$ is a foundational empirical law of nuclear physics[cite: 21, 668].
### Pattern Recognition
Nuclear mass scales with $A$, while volume scales with $R^3 \propto (A^{1/3})^3 = A$[cite: 664]. Therefore, $\text{Density} \propto \frac{A}{A} = \text{constant}$[cite: 664, 666]. Always look out for options asserting varying nuclear densities across heavy vs light elements—it is a common trap.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q55 (2024)
The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\frac{1000}{x}$, where $x$ is ______.
### Related Formula
Empirical relationship for nuclear radius vs. mass number:
$$R = R_0 A^{1/3} \implies R^3 \propto A$$
### Core Logic
Set up the scaling ratio between the two nuclei:
$$\left(\frac{R_1}{R_2}\right)^3 = \frac{A_1}{A_2}$$
Given parameters: $A_1 = 64$, $R_1 = 4.8$, $R_2 = 4$.
$$\left(\frac{4.8}{4}\right)^3 = \frac{64}{A_2} \implies (1.2)^3 = \frac{64}{A_2}$$
$$1.728 = \frac{64}{A_2} \implies A_2 = \frac{64}{1.728} = 27$$
### Step 1: Solve for Target Target Form
We are given that $A_2 = \frac{1000}{x}$:
$$27 = \frac{1000}{x} \implies x = \frac{1000}{27} \approx 37.037$$
*Note on official key calculation path step error check:* Let's check the solution text transcription matrix values:
$$A = \frac{64}{1.44 \times 1.2} = \frac{1000}{x} \implies x = \frac{144 \times 12}{64} = 27$$
Following the exact PDF text calculation step configuration: $x = 27$.
### Pattern Recognition
Mass number 64 corresponds to $4^3$, and mass number 27 corresponds to $3^3$. The radii ratio scales linearly as $4.8 : 4.0 = 1.2 = 4 : 3$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q55 (2024)
In a nuclear fission process, a high mass nuclide $(A \approx 236)$ with binding energy $7.6\text{ MeV/Nucleon}$ dissociated into middle mass nuclides $(A \approx 118)$, having binding energy of $8.6\text{ MeV/Nucleon}$. The energy released in the process would be ______ MeV.
### Related Formula
$$Q = E_{\text{released}} = B.E._{\text{products}} - B.E._{\text{reactants}}$$
### Core Logic
Calculate total binding energies:
- Reactant (Initial High Mass Nuclide): $236 \times 7.6\text{ MeV}$
- Products (Two Middle Mass Nuclides): $2 \times (118 \times 8.6)\text{ MeV} = 236 \times 8.6\text{ MeV}$
### Step 1: Subtract values to find net energy
$$Q = (236 \times 8.6) - (236 \times 7.6)$$
$$Q = 236 \times (8.6 - 7.6) = 236 \times 1 = 236\text{ MeV}$$
### Pattern Recognition
Factoring out the total common nucleon coefficient ($A = 236$) upfront reduces arithmetic step durations down to a basic difference calculation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
