Explanation

(c)
Radius of orbit ($r$)=$\frac{n^{2}h^{2}}{4\pi^{2}me^{2}} \times \frac{1}{Z}$
In it $h,\ \pi,\ m\ and\ e$ are constants, so after substituting these values, we get
$\ \ \ \ \ \ \ \ r = \frac{0.529n^{2}}{Z}\mathring{\mathrm{A}}$
$\ \ \ \ \ \ \ Z = 1\ for\ H$
$\therefore\ \ \ r_{H} = \frac{0.529n^{2}}{1}\mathring{\mathrm{A}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
The transition from $n = 2\ to\ n = 1$ in H-atom will have the same wavelength as the transition from $n = 4\ to\ n = 2$ in $He^{+}$ ion.