The ratio of radii of second orbit of hydrogen atom to fourth orbit of $\mathrm{He}^{+}$ion is
The ratio of radii of second orbit of hydrogen atom to fourth orbit of $\mathrm{He}^{+}$ion is
$1: 4$
$2: 1$
$1: 2$
$2: 3$
Solution
For hydrogen and hydrogen like atoms
$r_n=n^2 / Z$
where, $r_n$ is radius of $n$th orbit of revolution of electron, $Z$ is the atomic number.
For hydrogen atom, $Z=1$.
Hence, radius of second orbit is $r_2=\frac{(2)^2}{1}=4$
For helium atom, $Z=2$.
Hence, radius of fourth orbit is $r_4=\frac{(4)^2}{2}=8$
$\therefore$ Therefore, $\frac{r_2}{r_4}=\frac{4}{8}=\frac{1}{2} \Rightarrow 1: 2$