Question

Balmer gave an equation for wavelength of visible region of H-spectrum as $\lambda =\frac{K{n}^2}{n^2-4}$. Where n = principal quantum number of energy level, K = constant in terms of R (Rydberg constant).
The value of K in term of R is:

• A. $R$
• B. $\frac{R}{2}$
• C. $\frac{4}{R}$
• D. $\frac{5}{R}$

Answer 1

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

Given: $\lambda =\frac{K{n}^2}{n^2-4}...\left(i\right)$

Wave number $\stackrel{-}{v}$ for a Balmer transtiton $n\to 2$
$\stackrel{-}{v}=\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{n^2}]$
where $R=\text{Rydberg constant}$
$n=\text{Initial energy state}$
$\Rightarrow \lambda =\frac{4}{R}\frac{n^2}{n^2-4}..\left(ii\right)$

Comparing (i) and (ii),
$\Rightarrow k=\frac{4}{R}$