Question

If the ionization energy of $H{e}^{+}$ is $19.6\times {10}^{-10}J$ per atom then the energy of $B{e}^{3+}$ ion in the second stationary state is:

• A. $-4.9\times {10}^{-18}J$
• B. $-44.1\times {10}^{-18}J$
• C. $-11.025\times {10}^{-18}J$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

Ionization energy for an atom of atomic number $Z$ is

$(IE{)}_Z=(IE{)}_H\times \frac{Z^2}{n^2}$

Given, $(IE{)}_{H{e}^{+}}=19.6\times {10}^{-10}J/atom=(IE{)}_H\times \frac{2^2}{n^2}$------------------1

$(IE{)}_{B{e}^{3+}}=(IE{)}_H\times \frac{(4{)}^2}{n^2}$------------------2

$n=2,Z=4$

$\frac{Equation\left(2\right)}{Equation\left(1\right)}\Rightarrow \frac{(IE{)}_{B{e}^{3+}}}{(IE{)}_{H{e}^{+}}}=\frac{(4{)}^2}{(2{)}^2}\times \frac{2^2}{1}\times 19.6\times {10}^{-8}$

$(IE{)}_{B{e}^{3+}}=16\times 19.6\times {10}^{-8}=-313.6\times {10}^{-8}J$