COMEDK Exam 2019 Solved Paper

Excitation energy of hydrogen like ion in its first excitation state, ∆E=40.8‌eV
For the hydrogen like ion, energy in the ground state (n=1)
E=‌

−13.6Z2

n2

‌eV=−13.6Z2‌eV
Energy in the first excited state, E2‌=‌

−13.6Z2

n2

‌=‌

−13.6Z2

4

Excitation energy, ∆E=E2−E1
‌=(‌

−13.6Z2

4

+13.6Z2)‌eV
‌=(−3.4+13.6)Z2‌eV
⇒‌‌40.8‌eV‌=10.2Z2‌eV
⇒‌‌Z2‌=‌

40.8

10.2

⇒‌‌Z2‌=4
⇒‌‌Z‌=2
Thus, energy of the electron in the ground state is given
‌ as ‌‌E1=−13.6Z2‌eV
⇒E1=−13.6(2)2‌eV
⇒E1=−54.4‌eV
Hence, the energy needed to remove the electron from the ion in ground state (ionisation energy)
=−(−54.4)‌eV=54.4‌eV