Question

The ratio of the ionization energy of H atom and $B{e}^{+3}$ ion is :

• A. 1:1
• B. 1:3
• C. 1:9
• D. 1:16

Answer 1

Answer: (D)

1:16

Ionisation energy of H atom $=13.6eV$

Here, $B{e}^{3+}$ is H like atom.

$\therefore$ Ionisation energy of $B{e}^{3+}$ ion $=13.6\times Z^{2}eV$

$=13.6\times 4^{2}eV$

$=16\times 13.6eV$

Now, ratio of their ionisation energy:

$\frac{1E_H}{1E_{B{e}^{3+}}}=\frac{13.6eV}{16\times 13.6eV}=\frac{1}{16}=1:16$