Question

Balmer gives an equation for wavelength of visible radiation of H-spectrum as $\lambda =\frac{k{n}^2}{n^2-4}$. The value of $K$ in terms of Rydberg's constant, $R$ is

• A. $R$
• B. $4R$
• C. $R/4$
• D. $4/R$

Answer 1

The correct option is D $4/R$

Wavelength of visible radiation in Balmer series, $\frac{1}{\lambda }=R(\frac{1}{4}-\frac{1}{n^2})$ for $n\ge 2$

Rearranging we get, $\lambda =\frac{4}{R}.\frac{n^2}{n^2-4}=\frac{k{n}^2}{n^2-4}$

$⟹k=\frac{4}{R}$