Match the following items:A. Energy of ground | vedclass

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(C) For $He^+$ $(Z=2)$:$(A)$ Ground state energy $(n=1)$: $E_n = -13.6 	imes (Z^2/n^2) 	ext{ eV} = -13.6 	imes (4/1) = -54.4 	ext{ eV}$. Thus,$A-i

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$A-iv, B-ii, C-i, D-iii$

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(C) For $He^+$ $(Z=2)$:$(A)$ Ground state energy $(n=1)$: $E_n = -13.6 	imes (Z^2/n^2) 	ext{ eV} = -13.6 	imes (4/1) = -54.4 	ext{ eV}$. Thus,$A-iv$.$(B)$ Potential energy of $H$ atom $(Z=1)$ in $I$ orbit $(n=1)$: $PE = 2 	imes E_n = 2 	imes (-13.6 	ext{ eV}) = -27.2 	ext{ eV}$. Thus,$B-ii$.$(C)$ Kinetic energy of $II$ excited state $(n=3)$ of $He^+$: $KE = |E_n| = 13.6 	imes (Z^2/n^2) = 13.6 	imes (4/9) \approx 6.04 	ext{ eV}$. Thus,$C-i$.$(D)$ Ionization potential of $He^+$: Energy required to remove electron from $n=1$ to $\infty$,which is $54.4 	ext{ eV}$. Converting to Joules: $54.4 	imes 1.602 	imes 10^{-19} 	ext{ J} \approx 8.72 	imes 10^{-18} 	ext{ J}$. Thus,$D-iii$.Therefore,the correct match is $A-iv, B-ii, C-i, D-iii$.

$A$. Energy of ground state of $He^+$	$i$. $+6.04 \text{ eV}$
$B$. Potential energy of $I$ orbit of $H$ atom	$ii$. $-27.2 \text{ eV}$
$C$. Kinetic energy of $II$ excited state of $He^+$	$iii$. $8.72 \times 10^{-18} \text{ J}$
$D$. Ionization potential of $He^+$	$iv$. $-54.4 \text{ eV}$

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