Question

In hydrogen atom spectrum, (R → Rydberg's constant)
A. the maximum wavelength of the radiation of Lyman series is 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 9/R
D. the minimum wavelength of Lyman series is 5/4R
Choose the correct answer from the options given below :

• A. A, B and C Only
• B. A, B and D Only
• C. A, B Only
• D. B, D Only

Hint:

Minimum wavelength corresponds to the "series limit," which is always calculated by setting $n_2 = \infty$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ determines the wavelength of transitions. Maximum wavelength occurs for the smallest energy transition ($n_2 = n_1 + 1$), while minimum wavelength occurs for $n_2 = \infty$.

Step 2: Key Formula or Approach:
1. Lyman: $n_1 = 1$. Max $\lambda$ ($n_2=2$), Min $\lambda$ ($n_2=\infty$).
2. Paschen: $n_1 = 3$. Min $\lambda$ ($n_2=\infty$).
Step 3: Detailed Explanation:
Statement A: Lyman max $\lambda \implies \frac{1}{\lambda} = R(1 - \frac{1}{4}) = \frac{3R}{4} \implies \lambda = \frac{4}{3R}$. (Correct)
Statement B: Balmer series ($n_1=2$) falls in the visible range. (Correct)
Statement C: Paschen min $\lambda \implies \frac{1}{\lambda} = R(\frac{1}{9} - 0) = \frac{R}{9} \implies \lambda = \frac{9}{R}$. (Correct)
Statement D: Lyman min $\lambda \implies \frac{1}{\lambda} = R(1 - 0) = R \implies \lambda = \frac{1}{R}$. (Incorrect)
Step 4: Final Answer:
The correct statements are A, B and C Only.