Question

If the wavelength of the first line of the Lyman series for the hydrogen atom is 1216 Å, then the wavelength of the first line of the Balmer series of the hydrogen spectrum is –

• A. 1216 Å
• B. 6563 Å
• C. 912 Å
• D. 3648 Å

Answer 1

Answer: (B)

6563 Å

Answer 2

$\frac{1}{\lambda}$ = RZ² $\left( \frac{1}{n_{1}^{2}}–\frac{1}{n_{2}^{2}} \right)$