Question

If the ionisation energy of ${He}^{+}$ is $19.6\times {10}^{-18}J$ per atom then the energy of ${Be}^{3+}$ in the $2^{nd}$ stationary state is:

• A. $156.8\times {10}^{-17}J/atom$
• B. $15.68\times {10}^{-17}KJ/atom$
• C. $15.68\times {10}^{-17}J/atom$
• D. none of these

Answer 1

The correct option is D none of these

$E\alpha \frac{Z^2}{n^2}$

For ionization of $H{e}^{+}$

$Z=2,n=1$

$\therefore E=19.6\times {10}^{-18}J$

For $B{e}^{3+}$ in 2nd state

$Z=4$ $n=2$

$\therefore E^{\prime }=\frac{Z^2}{n^2}E$

$=\frac{16}{4}E=4E$

$=4\times 19.6\times {10}^{-18}$

$=7.84\times {10}^{-17}$