Atoms Previous Year Questions — JEE Main Physics
22 past-year Atoms questions from JEE Main (Physics).
Q11 (2025)
Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of $\mathrm{Li}^{++}$ ion in its ground state is given by $\frac{1}{\mathrm{X}}\mathrm{a}_0$ , where $\mathrm{X} = \_ \_ \_ .$
(Where $\mathbf{a}_0$ is the first Bohr's radius.)
- $2$
- $1$
- $3$
- $9$
### Related Formula
Bohr radius formula for hydrogen-like species:
$$r_n = a_0 \frac{n^2}{Z}$$
### Core Logic
For Lithium ion $\mathrm{Li}^{++}$:
- Atomic number $Z = 3$
- For ground state, the principal quantum number $n = 1$
Substitute $Z = 3$ and $n = 1$ into Bohr's radius formula:
$$r_1 = a_0 \frac{1^2}{3} = \frac{a_0}{3}$$
Comparing with the given expression $\frac{1}{X} a_0$:
$$\frac{1}{X} a_0 = \frac{a_0}{3} \implies X = 3$$
### Pattern Recognition
Sees: Ground state radius of hydrogen-like species.
Trap: Confusing the atomic number $Z$ of Lithium with Helium ($Z=2$) or Beryllium ($Z=4$).
Shortcut: Ground state radius of hydrogenic species is simply $a_0 / Z$. Since Lithium has $Z=3$, the ground state radius must be $a_0/3$ directly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Atoms
Q7 (2025)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only.
Reason R: The formulation of Bohr model does not include repulsive force between electrons.
In the light of the above statements, choose the correct answer from the options given below :
- Both A and R are true but R is NOT the correct explanation of A.
- A is false but R is true.
- Both A and R are true and R is the correct explanation of A.
- A is true but R is false.
### Related Formula
Bohr's electrostatic centripetal balance equation:
$$\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Z e^2}{r^2}$$
This basic equation matches ONLY a single electron orbiting a nucleus of charge $+Ze$.
### Core Logic
Assertion Analysis:
- Bohr's model matches single-electron species (such as $\mathrm{H}$, $\mathrm{He}^+$, $\mathrm{Li}^{2+}$, etc.). Hence, Assertion A is true.
Reason Analysis:
- The model is restricted because it models ONLY the attractive force between the positive nucleus and one orbiting electron. It cannot handle multi-electron systems due to the presence of inter-electron repulsive forces, which are not integrated into Bohr's simple formulation. Hence, Reason R is true.
Connection Check:
- The lack of repulsive forces is exactly why the model fails for multi-electron species and remains applicable only to single-electron (hydrogen-like) species. Thus, R is the correct explanation of A.
### Pattern Recognition
For single vs. multi-electron systems in atomic physics: Bohr model = single-electron ONLY. Quantum mechanics (Schrodinger) is required for multi-electron systems due to electron-electron interactions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Atoms
Q24 (2025)
An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{\text{th}}$ energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________.
### Related Formula
The energy of an electron in the $n$-th shell of a hydrogen atom is:
$$E_n = -\frac{13.6}{n^2}\mathrm{~eV}$$
The emitted photon energy during transition $n_i \rightarrow n_f$ is:
$$\Delta E = E_{n_i} - E_{n_f}$$$
### Core Logic
Given state:
- Initial state: "fourth excited state" $\Rightarrow n_i = 5$
- Emitted photon energy: $\Delta E = 2.86\mathrm{~eV}$
### Step 1: Calculate energy of initial state ($E_5$)
$$E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} = -0.544\mathrm{~eV}$$$
### Step 2: Calculate final state energy ($E_n$)
$$2.86 = E_5 - E_n = -0.544 - E_n$$
$$E_n = -0.544 - 2.86 = -3.404\mathrm{~eV}$$$
### Step 3: Determine the integer shell index ($n$)
$$-3.4 = -\frac{13.6}{n^2}$$
$$n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2$$
### Pattern Recognition
Memorizing the first few energy levels of the hydrogen atom saves calculation time:
- $E_1 = -13.6\mathrm{~eV}$
- $E_2 = -3.4\mathrm{~eV}$
- $E_3 = -1.51\mathrm{~eV}$
- $E_4 = -0.85\mathrm{~eV}$
- $E_5 = -0.54\mathrm{~eV}$
Recognizing that a transition from $-0.54\mathrm{~eV}$ by emitting $2.86\mathrm{~eV}$ lands exactly at $-3.4\mathrm{~eV}$ maps directly to $n=2$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q15 (2025)
In a hydrogen like ion, the energy difference between the $2^{\mathrm{nd}}$ excitation energy state and ground is $108.8\mathrm{eV}$ . The atomic number of the ion is
- 4
- 2
- 1
- 3
### Related Formula
For a hydrogen-like ion of atomic number $Z$, the transition energy between states $n_2$ and $n_1$ is:
$$\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \mathrm{~eV}$$$
### Core Logic
Identify the states:
- Ground state: $n_1 = 1$
- $2^{\mathrm{nd}}$ excitation energy state: $n_2 = 3$
Substitute values into the energy equation:
$$108.8 = 13.6 Z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$$
$$108.8 = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) = 13.6 Z^2 \times \frac{8}{9}$$
### Step 1: Solve for Z
Rearrange the equation to isolate $Z^2$:
$$Z^2 = \frac{108.8 \times 9}{13.6 \times 8}$$
Calculate the numerical value:
$$\frac{108.8}{13.6} = 8$$
$$Z^2 = \frac{8 \times 9}{8} = 9 \implies Z = 3$$
### Pattern Recognition
Sees: "$2^{\mathrm{nd}}$ excitation" $\implies n = 3$. Ground state $\implies n = 1$.
Shortcut: Energy transition ratio for $1 \to 3$ is always $\frac{8}{9}$ of $13.6 Z^2 \approx 12.1 Z^2$. Since $108.8 / 12.1 \approx 9$, $Z^2 = 9 \implies Z = 3$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Atoms
Q16 (2025)
For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.
- $5:36$
- $5:27$
- $3 : 4$
- $27:5$
### Related Formula
The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula:
$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$
To find the largest wavelength (minimum energy transition), select the adjacent higher shell $n_2 = n_1 + 1$.
### Core Logic
- **Lyman Series largest wavelength ($\lambda_L$)**: Transition from $n = 2 \to 1$
$$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$$
- **Balmer Series largest wavelength ($\lambda_B$)**: Transition from $n = 3 \to 2$
$$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}$$
### Step 1: Ratio Calculation
Now, compute the ratio of the wavelengths:
$$\frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$$$
### Pattern Recognition
Sees: Ratio of largest wavelengths of series.
Shortcut: The largest wavelength in a series starting at ground level $n_1$ is $\lambda \propto \frac{n_1^2 (n_1+1)^2}{2n_1 + 1}$.
For Lyman ($n_1=1$): $\lambda_L \propto \frac{4}{3}$.
For Balmer ($n_1=2$): $\lambda_B \propto \frac{36}{5}$.
Ratio: $\frac{4/3}{36/5} = \frac{5}{27}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
