Atoms Previous Year Questions — JEE Main Physics
22 past-year Atoms questions from JEE Main (Physics) — page 4.
Q53 (2024)
A electron of hydrogen atom on an excited state is having energy $E_n = -0.85 \mathrm{~eV}$. The maximum number of allowed transitions to lower energy level is ....
### Related Formula
$$E_n = -\frac{13.6}{n^2} \mathrm{~eV}$$
$$\text{Number of transitions} = \frac{n(n - 1)}{2}$$
### Core Logic
First, identify the principal quantum number $n$ corresponding to the energy $-0.85 \mathrm{~eV}$. Then, use the combinatorics formula to find the total possible downward emission transitions.
### Step 1: Find Quantum State
$$E_n = -\frac{13.6}{n^2}$$
$$-0.85 = -\frac{13.6}{n^2}$$
$$n^2 = \frac{13.6}{0.85} = 16$$
$n = 4$
### Step 2: Calculate Transitions
Maximum number of transitions from $n=4$ to lower levels ($n=3, 2, 1$):
$$= \frac{n(n - 1)}{2}$$
$$= \frac{4(4 - 1)}{2} = \frac{12}{2} = 6$$
### Pattern Recognition
Energy states in Hydrogen are heavily standardized: $n=1 \rightarrow -13.6$, $n=2 \rightarrow -3.4$, $n=3 \rightarrow -1.51$, $n=4 \rightarrow -0.85$. Recognize $-0.85 \mathrm{~eV}$ as state 4 immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q36 (2024)
If the wavelength of the first member of Lyman series of hydrogen is $\lambda$. The wavelength of the second member will be
- $\frac{27}{32} \lambda$
- $\frac{32}{27}\lambda$
- $\frac{27}{5} \lambda$
- $\frac{5}{27}\lambda$
### Related Formula
$$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$$
### Core Logic
For the first member of the Lyman series of hydrogen ($n_1 = 1, n_2 = 2$):
$$\frac{1}{\lambda} = \frac{13.6 Z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{2^2} \right]$$
$$\frac{1}{\lambda} = \frac{13.6 Z^2}{hc} \left[ \frac{3}{4} \right] \dots \dots (\text{i})$$
### Step 2: Second Member Calculation
For the second member of the Lyman series ($n_1 = 1, n_2 = 3$):
$$\frac{1}{\lambda'} = \frac{13.6 Z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right]$$
$$\frac{1}{\lambda'} = \frac{13.6 Z^2}{hc} \left[ \frac{8}{9} \right] \dots \dots (\text{ii})$$
### Step 3: Ratio
On dividing equation (i) by (ii):
$$\frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{3}{4} \times \frac{9}{8}$$
$$\frac{\lambda'}{\lambda} = \frac{27}{32}$$
$$\lambda' = \frac{27}{32} \lambda$$
### Pattern Recognition
Rydberg ratios between members of the same series are purely derived from the bracket terms $[1/n_1^2 - 1/n_2^2]$. For Lyman 1st and 2nd, the ratio is $(3/4) / (8/9) = 27/32$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q38 (2026)
If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m.
(Atomic number of gold = 79 and $\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9}$ in SI units)
- $2.95 \times 10^{-14}$
- $2.95 \times 10^{-16}$
- $3.85 \times 10^{-16}$
- $3.85 \times 10^{-14}$
### Related Formula
$$K_{\text{initial}} = U_{\text{closest approach}}$$
$$K = \frac{1}{4\pi \epsilon_{0}} \frac{(2e)(Ze)}{r_0}$$
### Core Logic
By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach ($r_0$).
$K_i + U_i = K_f + U_f$
$$K_i + 0 = 0 + \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{r_0}$$
### Step 1: Convert Energy and Solve
Initial kinetic energy $K_i = 7.7\text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19}\text{ J}$.
$$7.7 \times 10^6 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times (2 \times 1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19})}{r_0}$$
$$r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}$$
$$r_0 = \frac{9 \times 10^9 \times 158 \times 1.6 \times 10^{-19}}{7.7 \times 10^6}$$
$$r_0 = \frac{2275.2 \times 10^{-10}}{7.7 \times 10^6} = 295.48 \times 10^{-16}\text{ m} \approx 2.95 \times 10^{-14}\text{ m}$$
### Pattern Recognition
Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy $k(Z_1e)(Z_2e)/r_0$. Alpha particle has charge $2e$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q41 (2026)
The energy of an electron in an orbit of the Bohr's atom is $-0.04E_{0} \text{ eV}$ where $E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then $\frac{2\pi L}{h}$ is :
- $2$
- $4$
- $5$
- $6$
### Related Formula
$$E_n = \frac{E_0}{n^2}$$
$$L = \frac{nh}{2\pi}$$
### Core Logic
From Bohr's theory, the energy of an electron in the $n$-th orbit is inversely proportional to $n^2$:
$$E = -\frac{E_0}{n^2}$$
Equating this to the given energy:
$$-\frac{E_0}{n^2} = -0.04 E_0$$
$$\frac{1}{n^2} = 0.04 = \frac{1}{25}$$
$$n^2 = 25 \implies n = 5$$
### Step 1: Final Conclusion
Bohr's quantization condition for angular momentum:
$$L = \frac{nh}{2\pi}$$
Rearranging for the required term:
$$\frac{2\pi L}{h} = n$$
Since $n=5$, the value is $5$.
### Pattern Recognition
The expression $2\pi L / h$ is a direct request for the principal quantum number $n$. $E_n = E_1 / n^2$ allows finding $n$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q32 (2026)
The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ____ nm.
- $1875$
- $1550$
- $1217$
- $1784$
### Related Formula
$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$
### Core Logic
For smallest wavelength of Lyman series ($n_1 = 1, n_2 = \infty$):
$$\frac{1}{\lambda_{L, \text{min}}} = R \left(\frac{1}{1^2} - 0\right) \implies R = \frac{1}{91} \mathrm{~nm}^{-1}$$
For largest wavelength of Balmer series ($n_1 = 2, n_2 = 3$):
$$\frac{1}{\lambda_{B}} = R \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{1}{91} \times \frac{5}{36} \implies \lambda_B = \frac{91 \times 36}{5} = 655.2 \mathrm{~nm}$$
For largest wavelength of Paschen series ($n_1 = 3, n_2 = 4$):
$$\frac{1}{\lambda_{P}} = R \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{1}{91} \times \frac{7}{144} \implies \lambda_P = \frac{91 \times 144}{7} = 1872 \mathrm{~nm}$$
Difference in wavelengths:
$$\Delta \lambda = \lambda_P - \lambda_B = 1872 - 655.2 = 1216.8 \mathrm{~nm} \approx 1217 \mathrm{~nm}$$
### Step 1: Final Conclusion
The wavelength difference is approximately $1217 \mathrm{~nm}$.
### Pattern Recognition
Spectral lines: Smallest wavelength $\rightarrow n_2 = \infty$. Largest wavelength $\rightarrow n_2 = n_1 + 1$.
Rydberg constant $R = 1/\lambda_{L,\text{min}}$. Substitute values into Balmer ($2\rightarrow 3$) and Paschen ($3\rightarrow 4$) equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
