Question

Frequency of the series liit of Balmer series of hydrogen atom in terms of Rydberg's constant (R) and velocity of light (c) is

• A. 4Rc
• B. $\frac{4}{Rc}$
• C. Rc
• D. $\frac{Rc}{4}$

Hint:

Physics Tip: The series limit always corresponds to the shortest wavelength and highest frequency of that specific spectral series.

Answer 1

The Correct Option is D

Concept: Physics (Atoms) - Spectral Series of Hydrogen.

Step 1: State the Balmer series formula. The wavelength $\lambda$ for the Balmer series is given by the Rydberg formula: $$ \frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) \text{} $$ For the Balmer series, $n_1 = 2$.

Step 2: Determine the conditions for the series limit. The "series limit" refers to the transition from the highest possible energy level, where $n_2 = \infty$. $$ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \text{} $$ $$ \frac{1}{\lambda} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \text{} $$

Step 3: Calculate the frequency. Frequency $\nu$ is related to wavelength and the speed of light $c$ by the equation: $$ \nu = \frac{c}{\lambda} \text{} $$ Substituting $\frac{1}{\lambda} = \frac{R}{4}$: $$ \nu = c \times \frac{R}{4} = \frac{Rc}{4} \text{} $$ $$ \therefore \text{The frequency of the series limit is } \frac{Rc}{4}. \text{} $$