Consider one ${\mathrm{He}}^{+}$ ion in excited state (n = 5). Which of the following observations will hold true as per the Bohr's model

A multitude of ${\mathrm{He}}^{+}$ will produce 10 possible emission lines.

A single ${\mathrm{He}}^{+}$ can produce maximum 4 lines (not more)

$\begin{array}{l}5\stackrel{ }{\to }4\stackrel{ }{\to }3\stackrel{ }{\to }2\stackrel{ }{\to }1\\ {\mathrm{E}}_{\mathrm{n}}=-13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-13.6\times \frac{2^2}{5^2}\\ =-13.6\times \frac{4}{25}=-2.176\mathrm{eV}\\ \mathrm{IE}=+2.176\mathrm{eV}>2\mathrm{eV}\\ \frac{1}{\mathrm{\lambda }}={\mathrm{RZ}}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\\ \frac{1}{\mathrm{\lambda }}=\mathrm{R}.4\left[\frac{1}{4^2}-\frac{1}{5^2}\right]\\ \frac{1}{\mathrm{\lambda }}=4\mathrm{R}\left[\frac{9}{16\times 25}\right]\\ \frac{1}{\mathrm{\lambda }}=\frac{16\times 25}{4\times 9\times \mathrm{R}}=\frac{100}{9\mathrm{R}}>\frac{10}{\mathrm{R}}\\ {\mathrm{r}}_{\mathrm{n}}=0.53\times \frac{{\mathrm{n}}^2}{\mathrm{Z}}=0.53\times \frac{5^2}{\mathrm{Z}}\\ =6.625{\text{\hspace{0.17em}\hspace{0.17em}A}}^{\mathrm{o}}>6{\mathrm{A}}^{\mathrm{o}}\end{array}$