Q38 (2026)
In hydrogen atom spectrum, ($R \rightarrow$ Rydberg's constant)
A. the maximum wavelength of the radiation of Lyman series is $\frac{4}{3R}$
B. the Balmer series lies in the visible region of the spectrum
C. the minimum wavelength of the radiation of Paschen series is $\frac{9}{R}$
D. the minimum wavelength of Lyman series is $\frac{5}{4R}$
Choose the correct answer from the options given below:
- B, D Only
- A, B and C Only
- A, B and D Only
- A, B Only
### Related Formula
$$\frac{1}{\lambda} = R\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)$$
### Core Logic
Assess each statement using the Rydberg formula. Max wavelength corresponds to min energy (transition from $n_{1}+1 \to n_{1}$), and min wavelength corresponds to max energy ($n_{2} \to \infty$).
### Step 1: Evaluate A and D (Lyman Series)
For Lyman series, $n_{1} = 1$.
Maximum wavelength ($n_{2} = 2$):
$$\frac{1}{\lambda_{\max}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \implies \lambda_{\max} = \frac{4}{3R}$$
Statement A is correct.
Minimum wavelength ($n_{2} = \infty$):
$$\frac{1}{\lambda_{\min}} = R\left(1 - 0\right) = R \implies \lambda_{\min} = \frac{1}{R}$$
Statement D is incorrect.
### Step 2: Evaluate B (Balmer Series)
Balmer series corresponds to transitions to $n_{1} = 2$. These transitions primarily emit in the visible spectrum.
Statement B is correct.
### Step 3: Evaluate C (Paschen Series)
For Paschen series, $n_{1} = 3$.
Minimum wavelength ($n_{2} = \infty$):
$$\frac{1}{\lambda_{\min}} = R\left(\frac{1}{9} - 0\right) = \frac{R}{9} \implies \lambda_{\min} = \frac{9}{R}$$
Statement C is correct.
### Step 4: Final Conclusion
Statements A, B, and C are correct.
### Pattern Recognition
Sees: "minimum/maximum wavelength" → Max wavelength = adjacent orbital drop ($n+1$ to $n$). Min wavelength = drop from infinity ($\infty$ to $n$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
