Question

If the series limit wavelength of the Lyman series for hydrogen atom is 912 Å, then the series limit wavelength for the Balmer series for the hydrogen atom is –

• A. 912 Å
• B. 912 × 2 Å
• C. 912 × 4 Å
• D. $\frac{912}{2}$ Å

Answer 1

Answer: (C)

912 × 4 Å

Answer 2

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

$\frac{\frac{1}{912} = R \times Z^{2}\left( \frac{1}{1^{2}}–\frac{1}{\infty} \right)}{\frac{1}{\lambda} = R \times Z^{2}\left( \frac{1}{1^{2}}–\frac{1}{\infty} \right)}$