Question

Calculate the energy of electromagnetic radiation which is obtained for the largest wavelength in Balmer series $H{e}^{+}$ ion.

Answer 1

Rydberg's equation $\Rightarrow \frac{1}{\lambda }={Rz}^2(\frac{1}{n_1^2}-\frac{1}{n_1^2})$

Balmer series $\Rightarrow n_1=2$

${He}^{+}$ ion $\Rightarrow z=2$

Longest wavelength$\Rightarrow n_2=3$

$\frac{1}{\lambda }=R\left(z^2\right)[\frac{1}{2^2}-\frac{1}{3^2}]$

$\lambda =\frac{1}{R\left(4\right)}\left[\frac{4\times 9}{5}\right]$

$1641.1\stackrel{0}{A}$

longesxt wavelength if balmer series of ${He}^{+}$ion is $1641.1\stackrel{0}{A}$