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


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