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The ground state energy of hydrogen atom is $-13.6$ eV. The energy needed to ionize hydrogen atom from its second excited state will be:
A
13.6 eV
B
6.8 eV
C
1.51 eV
D
3.4 eV
Detailed Solution
$E_n=\dfrac{-13.6Z^2}{n^2}$ eV. Ground state: $n=1$; second excited state: $n=3$. $\dfrac{E_3}{E_1}=\left(\dfrac{n_1}{n_3}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}$. $E_3=\dfrac{E_1}{9}=\dfrac{13.6\ eV}{9}=1.51$ eV. Ionization energy from the second excited state equals $|E_3|=1.51$ eV.