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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For $Li^{+}$,the

Vedclass · Accepted Answer

$4.905 	imes 10^{-18} \ J, 0.0705 \ nm$

Vedclass · Answer

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For $Li^{+}$,the atomic number $Z = 3$ and for the second orbit,$n = 2$.$E = -13.6 	imes \frac{3^2}{2^2} = -13.6 	imes \frac{9}{4} = -30.6 \ eV$.Converting to Joules: $E = -30.6 	imes 1.602 	imes 10^{-19} \ J \approx -4.905 	imes 10^{-18} \ J$.The magnitude of energy is $4.905 	imes 10^{-18} \ J$.The radius of the $n^{th}$ orbit is given by $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.$r_2 = 0.529 	imes \frac{2^2}{3} = 0.529 	imes \frac{4}{3} = 0.7053 \ \mathring{A}$.Since $1 \ \mathring{A} = 0.1 \ nm$,$r_2 = 0.07053 \ nm \approx 0.0705 \ nm$.Thus,the correct option is $(D)$.

Calculate the energy associated with the second orbit of $Li^{+}$ and find its radius.

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