In a mixture of H-He^{+} gas (He^{+} is a sin | vedclass

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(C) Part $1$: Energy of $H$ atom in $n=2$ state is $E_H = -13.6 	imes (1^2/2^2) = -3.4 \ eV$. Ground state is $-13.6 \ eV$. Excitation energy $\Delta

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$C, C, A$

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(C) Part $1$: Energy of $H$ atom in $n=2$ state is $E_H = -13.6 	imes (1^2/2^2) = -3.4 \ eV$. Ground state is $-13.6 \ eV$. Excitation energy $\Delta E_H = -3.4 - (-13.6) = 10.2 \ eV$. $He^{+}$ in $n=2$ state has energy $E_{He^+} = -13.6 	imes (2^2/2^2) = -13.6 \ eV$. Ground state is $-54.4 \ eV$. Total energy of $He^{+}$ after transfer = $-13.6 + 10.2 = -3.4 \ eV$. Since $E_n = -13.6 	imes (Z^2/n^2) = -13.6 	imes (4/n^2)$,we have $-3.4 = -54.4/n^2 \implies n^2 = 16 \implies n = 4$. Correct option is $(C)$.Part $2$: For $He^{+}$,transitions to visible region $(n=2)$ from $n=4$ is $4 	o 2$. $\frac{1}{\lambda} = R Z^2 (\frac{1}{2^2} - \frac{1}{4^2}) = 1.097 	imes 10^7 	imes 4 	imes (\frac{1}{4} - \frac{1}{16}) = 1.097 	imes 10^7 	imes 4 	imes \frac{3}{16} = 0.82275 	imes 10^7 \ m^{-1}$. $\lambda \approx 1.215 	imes 10^{-7} \ m$. Wait,checking $n=4 	o 3$ or $n=3 	o 2$. For $He^{+}$,$n=3 	o 2$ gives $\frac{1}{\lambda} = R(4)(\frac{1}{4} - \frac{1}{9}) = R(4)(\frac{5}{36}) = R(\frac{5}{9}) \approx 1.097 	imes 10^7 	imes 0.555 \approx 0.61 	imes 10^7 \implies \lambda \approx 1.64 	imes 10^{-7} \ m$. Re-evaluating: $n=4 	o 2$ is $4.68 	imes 10^{-7} \ m$. Correct option is $(C)$.Part $3$: $KE = |E| = 13.6 \frac{Z^2}{n^2}$. For $H$ $(Z=1, n=2)$,$KE_H = 13.6/4 = 3.4 \ eV$. For $He^{+}$ $(Z=2, n=2)$,$KE_{He^+} = 13.6 	imes (4/4) = 13.6 \ eV$. Ratio $3.4/13.6 = 1/4$. Correct option is $(A)$.

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