Answer
\( \mathrm{X} = \) carbon and \( m = 5 \)
Work Step by Step
We have:
$E_n=-\frac{R_HZ^2}{n^2}$
for one-electron species.
The ionization energy for the transition from $n=1$ to $n=\infty$ is:
$E_{\infty}-E_1=-E_1=\frac{R_HZ^2}{1^2}=R_HZ^2$
We solve for $Z$:
$4.72\times 10^4\times \frac{1}{6.022\times 10^{23}}\times 1000=2.178\times 10^{-18}Z^2$
$Z=6$
The element with $Z=6$ is carbon (C), so $X=C$ and the charge for a one-electron carbon ion is $5^+$ ($m=5$). The one-electron ion is $C^{5+}$.
