For wavelength of visible radiation of hydrogen spectrum Balmer gave an equation as…

For wavelength of visible radiation of hydrogen spectrum Balmer gave an equation as $\lambda=\frac{\left(\mathrm{km}^2\right)}{\left(\mathrm{m}^2-4\right)}$, where $m$ is the integer value. The value of $k$ in terms of Rydberg's constant $R$ is
  1. $\frac{R}{4}$
  2. $\frac{4}{R}$
  3. $R$
  4. $4 R$

Solution

Wavelength of visible radiation in Balmer series is given by, $\begin{aligned} & \frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{m^2}\right) \text { for } m \geq 2 \\ & \Rightarrow \lambda=\frac{4}{R} \cdot \frac{m^2}{m^2-4}=\frac{k m^2}{m^2-4}\end{aligned}$ $\therefore k=\frac{4}{R}$

Asked in: MHT CET 2022 (06 Aug Shift 2)

Practice more Atomic Physics questions on Aicharya