Question

The excitation energy of a hydrogen like ion to first excited state is 40.8 eV. The energy needed to remove the electron from the ion the ground state is

• A. 54.4 eV
• B. 62.6 eV
• C. 72.6 eV
• D. 58.6 eV.

Answer 1

The correct option is A 54.4 eV

For a hydrogen atom like system the energy for $n^{th}$ level is given by $E_n=\frac{-13.6Z^2}{n^2}eV$

So the ground state energy will be $E_1=\frac{-13.6Z^2}{1}eV$

and for the state $n=2$ the energy is $E_2=\frac{-13.6Z^2}{4}eV$

So the excitation energy foe this state will be $E_2-E_1=10.2Z^{2}eV$

as it is given to be $40.8eV$ so $Z^2=4$ or $Z=2$

So the energy in ground state will be $E_1=-54.4eV$

so the inonization energy will be $-(-54.4eV)=54.4eV$