Ionisation energy of He^{+} is 19.6 times 10^ | vedclass

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(B) Ionisation energy $(IE)$ is the energy required to move an electron from the ground state to infinity.$IE = E_{\infty} - E_{1} = 0 - E_{1} = -E_{1

Vedclass · Accepted Answer

$-4.41 	imes 10^{-17} \, J \, atom^{-1}$

Vedclass · Answer

(B) Ionisation energy $(IE)$ is the energy required to move an electron from the ground state to infinity.$IE = E_{\infty} - E_{1} = 0 - E_{1} = -E_{1}$Therefore,$E_{1}$ of $He^{+} = -19.6 	imes 10^{-18} \, J \, atom^{-1}$.Energy of a hydrogen-like species at state $n$ is given by $(E_{n})_{species} = (E_{1})_{H} 	imes \frac{Z^{2}}{n^{2}}$.For $He^{+}$ $(Z=2, n=1)$: $(E_{1})_{He^{+}} = (E_{1})_{H} 	imes 2^{2} = -19.6 	imes 10^{-18} \, J \, atom^{-1}$.So,$(E_{1})_{H} = \frac{-19.6 	imes 10^{-18}}{4} \, J \, atom^{-1}$.For $Li^{2+}$ $(Z=3, n=1)$: $(E_{1})_{Li^{2+}} = (E_{1})_{H} 	imes 3^{2} = \frac{-19.6 	imes 10^{-18}}{4} 	imes 9 = -4.41 	imes 10^{-17} \, J \, atom^{-1}$.

Ionisation energy of $He^{+}$ is $19.6 \times 10^{-18} \, J \, atom^{-1}$. The energy of the first stationary state $(n = 1)$ of $Li^{2+}$ is

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