Atoms And Nuclei Question 274
Question: As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z=3) is
Options:
A) 1.51
B) 13.6
C) 40.8
D) 122.4
Show Answer
Answer:
Correct Answer: D
Solution:
- $ E_{n}=-13.6\frac{{{( Z )}^{2}}}{( n^{2} )}eV $ Therefore, ground state energy of double ionized lithium atom (Z=3, n=1) will be $ E_{1}=( -13.6 )\frac{{{( 3 )}^{2}}}{{{( 1 )}^{2}}}=-122.4eV $
$ \therefore $ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4eV
BETA
