Question

Taking Rydberg’s constant $R_{H} = 1.097 \times 10^{7}m$ first and second wavelength of Balmer series in hydrogen spectrum is.

• A. 2000 Å, 3000 Å
• B. 1575 Å, 2960 Å
• C. 6529 Å, 4280 Å
• D. 6552 Å, 4863 Å

Answer 1

Answer: (D)

6552 Å, 4863 Å

Answer 2

$\frac{1}{\lambda} = R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack.$ For first wavelength, $n_{1} = 2$, $n_{2} = 3$ ⇒$\lambda_{1} = 6563Å$. For second wavelength, $n_{1} = 2$, $n_{2} = 4$

⇒ $\lambda_{2} = 4861Å$