Question

The ionization energy of the electron in the lowest orbit of hydrogen atom is $13.6eV$. The energies required in $eV$ to remove electron from three lowest orbits of hydrogen atom are

• A. 13.6, 6.8, 8.4
• B. 13.6, 10.2, 3.4
• C. 13.6, 27.2, 40.8
• D. 13.6, 3.4, 1.51

Answer 1

Answer: (D)

$I.E.=E_{\infty }-E_1$
$I.E.=13.6eV$ given
$E_{\infty }=0$
$13.6=0-E_1$
$E_1=-13.6eV$
$E_2=\frac{E_1}{n^2}\times (1{)}^2=-\frac{13.6}{4}=-3.4eV$
$E_3=\frac{E_1}{(3{)}^2}=\frac{-13.6}{9}=-1.51eV$
$\therefore I\cdot E_1=E_{\infty }-E_1=0-(-13.6)=13.6eV$
$I\cdot E_2=E_{\infty }-\left(E_2\right)=0-(-3.4)=3.4eV$
$I\cdot E_3=E_{\infty }-E_3=0(-(-1.51eV))=1.51eV$