Question

As per the Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of a double ionized Li atom (Z = 3) is

Answer 1

Explanation:
Energy of the nth orbit by Bohr was given by:
En=RH×Z2n2eV
where,E= energyRH= Rydberg's ConstantZ= atomic number =3 (for lithium)n= number of orbit.
Putting the values, in above equation, we get
Energy of the first shell(n=1) in hydrogen atom:
z=1−13.6eV=RH×12122RH=−13.6eV
To find energy value of electron in the excited state of Li2+ is:
Li:1s22s1Li2+:1s1Z=3,n=1En=−13.6×3212 eV=−122.4eV
Hence, the correct answer is -122.4.