Question

The energy of hydrogen atom in ground state is $13.6eV$. The ionization energy of singly ionized helium $H{e}^{2+}$ will be

• A. 13.6 eV
• B. 3.4 eV
• C. 27.2 eV
• D. 54.4 eV

Answer 1

Answer: (D)

The energy level of H-like atoms are given by
$E_n=\frac{Z^{2}Rhc}{n^2}$
$=\frac{-13.6Z^2}{n^2}eV$
For ionisation, the transition is from $n=1$ to $n=\infty$
$E_{\text{ionisation }}=E_{\infty }-E_1$
$=-13.6Z^2\left(\frac{1}{{\infty }^2}-\frac{1}{1^2}\right)$
$=13.6Z^{2}eV$
For hydrogen
$Z=1$
$\therefore {\left(E_{\text{ion }}\right)}_n=13.6eV...(i)$
For singly ionised helium $(Z=2)$
${\left(E_{\text{ion }}\right)}_{He}=13.6\times (2{)}^{2}eV$
$\therefore \frac{{\left(E_{\text{ion }}\right)}_{He}}{{\left(E_{\text{ion }}\right)}_H}=\frac{13.6\times 2^2}{13.6}=4$
$\Rightarrow {\left(E_{\text{ion }}\right)}_{He}=4\times 13.6eV$[from Eq. (i)]
$=54.4eV$