Question

The ratio of maximum to minimum wavelength in Balmer series of hydrogen atom is

• A. $36 : 5$
• B. $3 : 4$
• C. $9 : 5$
• D. $5 : 9$

Hint:

You can find this ratio directly without full fraction inversions by dividing the brackets inside the Rydberg setup: $\frac{\lambda_{max}}{\lambda_{min}} = \frac{(1/4 - 1/\infty)}{(1/4 - 1/9)} = \frac{1/4}{5/36} = \frac{36}{20} = \frac{9}{5}$. This layout saves several algebra transformation steps.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the ratio of the longest (maximum) wavelength $\lambda_{max}$ to the shortest (minimum) wavelength $\lambda_{min}$ emitted within the Balmer spectral line series of a hydrogen atom.

Step 2: Key Formula or Approach:
The Rydberg formula for the wavelength of spectral lines emitted by hydrogen atom electronic transitions is:
$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ For the Balmer series, the lower energy terminal state is always $n_1 = 2$.
1. The maximum wavelength ($\lambda_{max}$) corresponds to the lowest energy jump, which comes from the adjacent level: $n_2 = 3$.
2. The minimum wavelength ($\lambda_{min}$) corresponds to the highest possible energy jump, which comes from infinity: $n_2 = \infty$.

Step 3: Detailed Explanation:
Let's first determine the expression for $\lambda_{max}$ by setting $n_1 = 2$ and $n_2 = 3$:
$$\frac{1}{\lambda_{max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36}$$ Inverting this fraction gives:
$$\lambda_{max} = \frac{36}{5R}$$ Next, determine the expression for $\lambda_{min}$ by setting $n_1 = 2$ and $n_2 = \infty$:
$$\frac{1}{\lambda_{min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$$ Inverting this fraction gives:
$$\lambda_{min} = \frac{4}{R}$$ Now, calculate the requested ratio of maximum to minimum wavelength:
$$\frac{\lambda_{max}}{\lambda_{min}} = \frac{\frac{36}{5R}}{\frac{4}{R}} = \frac{36}{5R} \times \frac{R}{4} = \frac{36}{20} = \frac{9}{5}$$

Step 4: Final Answer:
The ratio of the maximum to minimum wavelength in the Balmer series is $9 : 5$, matching option (C).