The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4861 Å. The wavelength of the first line is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → ~ 6562 Å
The Balmer lines end on n = 2. The first line comes from n = 3 and the second line from n = 4; by the Rydberg formula
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right) \)
\( \displaystyle \text{First line } (n = 3)\text{: } \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \)
\( \displaystyle \text{Second line } (n = 4)\text{: } \frac{1}{4} - \frac{1}{16} = \frac{3}{16} \)
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{3/16}{5/36} = \frac{27}{20} \)
\( \displaystyle \lambda_1 = \frac{27}{20} \times 4861 \)
So \( \lambda_1 \approx 6562 \) Å.