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The photon emitted due to electronic transition from 5th excited state to 2nd excited state in $L i^{2 +}$ is used to excite $H e^{+}$ already in first excited state. $H e^{+}$ ion after absorbing the photon fully reaches an orbit having kinetic energy equal to____ eV ${atom}^{- 1}$
Given $E_{n} = \frac{- 13.6 \times Z^{2}}{n^{2}}$ eV ${atom}^{- 1}$

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answer is 3.4.

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Detailed Solution

$L i^{+ 2} H e^{+} R_{H} 3^{2} (\frac{1}{3^{2}} - \frac{1}{6^{2}}) = R_{H} 2^{2} (\frac{1}{2^{2}} - \frac{1}{n_{2}^{2}}) (1 - \frac{1}{2^{2}}) = 1 - \frac{2^{2}}{n_{2}^{2}} n_{2} = 4 E = \frac{- 13.6 \times 2^{2}}{4^{2}} = - 3.4 e V K E = + 3.4 e V$

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