AP EAPCET 18 May 2024 Shift 1 Solved Paper

Given:
The binding energy for a hydrogen atom is 13.6 eV .
We know:
The energy of an electron in the n-th state is given by:
En=−13.6‌

Z2

n2

For Li2+, the atomic number Z=3.
To find the energy required to remove the electron from the first excited state (n=2) :
Substitute into the equation:
‌E2=−13.6×‌

32

22

=−13.6×‌

9

4

‌E2=−30.6eV
Since E2+E=0, we have:
−30.6+E=0
Thus, solving for E :
E=+30.6eV
Therefore, the energy required to remove the electron from the first excited state of Li2+ is 30.6 eV .