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. $\frac{R}{4}$
• D. $\frac{4}{R}$

Answer 1

The correct option is D $\frac{4}{R}$

Generally the wavelength for Balmer series is given by, $\frac{1}{\lambda }=R(1/2^2-1/n^2)$

Now substituting the value of $\lambda$ given in above expression, $\frac{n^2-4}{k{n}^2}=R(1/2^2-1/n^2)$

or $\frac{n^2-4}{k{n}^2}=R\left[\frac{n^2-4}{4n^2}\right]$

or $k=\frac{4}{R}$