Question

The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to:

• A. $n_f = 2$ and $n_i = 3$
• B. $n_f = 3$ and $n_i = 4$
• C. $n_f = 2$ and $n_i = 4$
• D. $n_f = 2$ and $n_i = \infty$

Hint:

For the Balmer series, transitions end at $n_f = 2$. The spectral lines are numbered based on the value of $n_i$ (e.g., $n_i = 3$ for the first line, $n_i = 4$ for the second line).

Answer 1

The Correct Option is C

The Balmer series encompasses transitions terminating at the energy level $n_f = 2$. The second spectral line within this series originates from a transition where the initial energy level is $n_i = 4$ and the final energy level is $n_f = 2$.The wavelength of the emitted radiation is determined by the formula:\[\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right),\]where $R_H$ represents the Rydberg constant. For the second spectral line, the parameters are $n_f = 2$ and $n_i = 4$.Substituting these values into the formula yields:\[\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right).\]Simplification of the expression results in:\[\frac{1}{\lambda} = R_H \cdot \frac{4 - 1}{16} = R_H \cdot \frac{3}{16}.\]Consequently, the definitive transition corresponding to the second line of the Balmer series is:\[\boxed{n_f = 2 \text{ and } n_i = 4}.\]