According to Bohr's theory the energy required for an electron in the Li 2+ ion to be emitted from n=2 state is (given that the ground state ionization energy of hydrogen atom is 13.6 eV )

Question

According to Bohr's theory the energy required for an electron in the Li 2+ ion to be emitted from n=2 state is (given that the ground state ionization energy of hydrogen atom is 13.6 eV ) (A) 61.2 eV (B) 13.6 eV (C) 30.6 eV (D) 10.2 eV

Tardigrade Admin · Accepted Answer

Energy of electron in 2 text nd orbit of Li +2=-13.6 (Z2/n2) =(-13.6 ×(3)2/(2)2)=-30.6 eV Energy required =0-(-30.6)=30.6 eV

Q. According to Bohr's theory the energy required for an electron in the $L i^{2 +}$ ion to be emitted from $n = 2$ state is (given that the ground state ionization energy of hydrogen atom is $13.6 e V)$

61.2 eV

13.6 eV

30.6 eV

10.2 eV

Energy of electron in $2^{nd}$ orbit of $L i^{+ 2} = - 13.6 \frac{Z ^{2}}{n ^{2}}$
$= \frac{- 13.6 \times ( 3 ) ^{2}}{( 2 ) ^{2}} = - 30.6 e V$
Energy required $= 0 - (- 30.6) = 30.6 e V$

Q. According to Bohr's theory the energy required for an electron in the Li2+ ion to be emitted from n=2 state is (given that the ground state ionization energy of hydrogen atom is 13.6eV)