Atoms Previous Year Questions — JEE Main Physics
22 past-year Atoms questions from JEE Main (Physics) — page 2.
Q20 (2025)
The number of spectral lines emitted by atomic hydrogen that is in the $4^{\text{th}}$ energy level, is:
- $6$
- $0$
- $3$
- $1$
### Related Formula
$$N = \frac{n(n - 1)}{2}$$
where $n$ is the principal quantum number of the starting energy level.
### Core Logic
For a hydrogen sample initially in the $n = 4$ level, the possible downward transition pathways to reach the ground state ($n=1$) are:
{{SOL_IMG1}}
Using the combination formula for all transitions:
$$N = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = 6$$
Thus, 6 distinct spectral lines are generated.
### Pattern Recognition
Think of it as counting combinations of transitions between levels: $_nC_2$. For $n=4$, $_4C_2 = 6$ lines.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q5 (2025)
During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-
- 3000 Å
- 6000 Å
- 4000 Å
- 2000 Å
### Related Formula
The energy of the emitted photon during an atomic transition between energy states is given by:
$$\Delta E = \frac{hc}{\lambda} $$
where $h$ is Planck's constant, $c$ is speed of light, and $\lambda$ is the photon wavelength.
### Core Logic
Write equations for the energy transitions from the layout of levels {{SOL_IMG1}} :
$$E_{A} - E_{C} = \frac{hc}{\lambda_{AC}} = \frac{hc}{2000\text{ \AA}} \quad \dots (i)$$
$$E_{B} - E_{C} = \frac{hc}{\lambda_{BC}} = \frac{hc}{6000\text{ \AA}} \quad \dots (ii)$$
### Step 1: Finding Transition A to B
Subtracting equation $(ii)$ from equation $(i)$ gives the net transition energy from A to B :
$$E_{A} - E_{B} = (E_{A} - E_{C}) - (E_{B} - E_{C})$$
$$\frac{hc}{\lambda_{AB}} = \frac{hc}{2000} - \frac{hc}{6000}$$
$$\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}$$
$$\lambda_{AB} = 3000\text{ \AA}$$
### Pattern Recognition
Energy differences add linearly, which means their corresponding inverse wavelengths satisfy a parallel reciprocal subtraction rule: $\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q7 (2025)
The frequency of revolution of the electron in Bohr's orbit varies with $n$ , the principal quantum number as
- $\frac{1}{n}$
- $\frac{1}{\mathrm{n}^3}$
- $\frac{1}{\mathrm{n}^4}$
- $\frac{1}{n^2}$
### Related Formula
The orbital frequency of revolution $f$ of an electron is inversely proportional to its time period $T$:
$$f = \frac{1}{T} = \frac{v}{2\pi r}$$
In Bohr's Atomic Model:
* Velocity $v \propto \frac{Z}{n}$
* Radius $r \propto \frac{n^2}{Z}$
### Core Logic
Substitute the proportional relationships of $v$ and $r$ into the frequency expression:
$$f \propto \frac{\left(\frac{1}{n}\right)}{n^2} \implies f \propto \frac{1}{n^3}$$
### Step 1: Verification
Thus, the frequency varies inversely with the cube of the principal quantum number: $f \propto \frac{1}{n^3}$.
### Pattern Recognition
Remember the sequence of powers of $n$ in Bohr's model: radius expands as $n^2$, velocity drops as $n^{-1}$, angular momentum grows as $n^1$, and orbital time period or frequency changes as $n^3$ or $n^{-3}$ respectively.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q49 (2024)
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:
- 1.5 eV
- 13.6 eV
- 1.9 eV
- 12.1 eV
### Related Formula
Bohr state energy level values:
$$E_n = -\frac{13.6}{n^2}\mathrm{~eV}$$
Transition excitation requirement:
$$\Delta E = E_{\text{final}} - E_{\text{initial}}$$
### Core Logic
To emit radiation in the Balmer series, the hydrogen electron must first be excited to at least the $n=3$ shell. This allows it to jump down to $n=2$ and produce the first spectral line of the Balmer series.
Energy required to transition from ground state ($n=1$) to $n=3$:
$$E_1 = -13.6\mathrm{~eV}$$
$$E_3 = -\frac{13.6}{3^2} = -1.51\mathrm{~eV}$$
### Step 1: Calculate Energy Gap
$$\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\mathrm{~eV} \approx 12.1\mathrm{~eV}$$
### Pattern Recognition
Balmer emissions always return down to $n=2$. Thus, the initial excitation starting from ground state $n=1$ must reach at least $n=3$ to create a valid Balmer transition.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q49 (2024)
Given below are two statements:
Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model.
Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model.
In the light of the above statements, choose the most appropriate from the options given below:
- Both statement I and statement II are false.
- Statement I is false but statement II is true
- Statement I is true but statement II is false.
- Both statement I and statement II are true.
### Core Logic
* **Statement I**: According to Rutherford's planetary model of the atom, almost the entire mass of an atom and all of its positive charge are concentrated in a centrally located small volume called the nucleus, with electrons revolving around it. Thus, Statement I is **true**.
* **Statement II**: Thomson's model of the atom (plum-pudding model) describes the atom as a spherical cloud of positive charge with electrons embedded in it. This model was proposed before Rutherford's nuclear model and is not a special case of Rutherford's model. Thus, Statement II is **false**.
### Step 1: Conclusion
Therefore, Statement I is true but Statement II is false.
### Pattern Recognition
Understand the evolutionary timeline of atomic models:
1. Thomson's Plum Pudding Model (spherical cloud with embedded electrons)
2. Rutherford's Nuclear Model (planetary orbital model with tiny centralized nucleus)
They are conceptually distinct, making Statement II false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
