Question

The ionization energy of the electron in the lowest orbit of hydrogen atom is $13.6\, eV$. 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)

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