An excited He^+ ion emits two photons in succ | vedclass

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(C) The energy of a photon is given by $E = \frac{1240}{\lambda} \, eV$. For the first transition, $\lambda_1 = 108.5 \, nm$, so $E_1 = \frac{1240}{10

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$5$

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(C) The energy of a photon is given by $E = \frac{1240}{\lambda} \, eV$. For the first transition, $\lambda_1 = 108.5 \, nm$, so $E_1 = \frac{1240}{108.5} \approx 11.43 \, eV$. For the second transition, $\lambda_2 = 30.4 \, nm$, so $E_2 = \frac{1240}{30.4} \approx 40.79 \, eV$. The total energy released is $E_{total} = E_1 + E_2 = 11.43 + 40.79 = 52.22 \, eV$. The energy of an electron in the $n$-th state of a hydrogen-like ion is $E_n = -13.6 \, Z^2 \frac{1}{n^2}$. For $He^+$, $Z = 2$, so $E_n = -13.6 	imes 4 	imes \frac{1}{n^2} = -54.4 \frac{1}{n^2} \, eV$. The transition is from state $n$ to ground state $(n=1)$, so $E_{total} = E_n - E_1 = -54.4 \left(\frac{1}{n^2} - 1ight) = 54.4 \left(1 - \frac{1}{n^2}ight)$. Setting $54.4 \left(1 - \frac{1}{n^2}ight) = 52.22$, we get $1 - \frac{1}{n^2} = \frac{52.22}{54.4} \approx 0.96$. $\frac{1}{n^2} = 1 - 0.96 = 0.04$. $n^2 = \frac{1}{0.04} = 25$, so $n = 5$.

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