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Find the quantum no 'n' corresponding to the excited state of $He^+$ ion if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4nm


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Given :
$\lambda_2=30.4\times 10^{-7}cm$
$\lambda_1=108.5\times 10^{-7}cm$
Let excited state of $He^+$ be $n_2$.It comes from $n_2$ to $n_1$ and then $n_1$ to 1 to emit two successive photon.
$\large\frac{1}{30.4\times 10^{-7}}=$$109678\times 4\big[\large\frac{1}{1^2}-\frac{1}{n_1^2}\big]$
$\therefore n_1=2$
Now for $\lambda_1:n_1=2$
$\large\frac{1}{108.5\times 10^{-7}}=$$109678\times 4\big[\large\frac{1}{2^2}-\frac{1}{n_2^2}\big]$
$\therefore$ excited state of He is $5^{th}$ orbit.
Hence (d) is the correct answer.
answered Jan 16, 2014 by sreemathi.v

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