For He^{+}, a transition takes place from the | vedclass

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(B) For a single electron system, the radius of the $n^{th}$ orbit is given by $r_n = a_0 	imes \frac{n^2}{Z}$.Given $Z=2$ for $He^{+}$.For the initi

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$30$

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(B) For a single electron system, the radius of the $n^{th}$ orbit is given by $r_n = a_0 	imes \frac{n^2}{Z}$.Given $Z=2$ for $He^{+}$.For the initial orbit, $r_2 = 105.8 \ pm$:$105.8 = 52.9 	imes \frac{n_2^2}{2} \implies n_2^2 = \frac{105.8 	imes 2}{52.9} = 4 \implies n_2 = 2$.For the final orbit, $r_1 = 26.45 \ pm$:$26.45 = 52.9 	imes \frac{n_1^2}{2} \implies n_1^2 = \frac{26.45 	imes 2}{52.9} = 1 \implies n_1 = 1$.The transition is from $n=2$ to $n=1$.The energy of the emitted photon is $\Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.$\Delta E = 2.2 	imes 10^{-18} 	imes (2)^2 	imes \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 2.2 	imes 10^{-18} 	imes 4 	imes \left( 1 - 0.25 ight) = 8.8 	imes 10^{-18} 	imes 0.75 = 6.6 	imes 10^{-18} \ J$.Using $\Delta E = \frac{hc}{\lambda}$, we get $\lambda = \frac{hc}{\Delta E} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{6.6 	imes 10^{-18}} = 3 	imes 10^{-8} \ m = 30 \ nm$.

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