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

Correct Answer: -122.4

Explanation:
Energy of the nth orbit by Bohr was given by:
$E_n=R_H\times \frac{Z^2}{n^2}eV$
where,$E=$ energy$R_H=$ Rydberg's Constant$Z=$ 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$$-13.6eV=R_H\times {\frac{1^2}{1^2}}^2$$R_H=-13.6eV$
To find energy value of electron in the excited state of $L{i}^{2+}$ is:
$Li:1s^{2}2s^1$$L{i}^{2+}:1s^1$$Z=3,n=1$$E_n=-13.6\times \frac{3^2}{12}eV=-122.4eV$
Hence, the correct answer is -122.4.