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An excited $He^{+}$ ion emits two photons in succession, with wavelengths $108.5 nm$ and $30.4 nm$ , in making a transition to ground state. The quantum number $n$ , corresponding to its intial excited state is (for photon of wavelength $\lambda$ , energy $E=\frac{124eV}{\lambda}$ ):

Option 1)
$n=5$

Option 2)
$n=6$

Option 3)
$n=7$

Option 4)
$n=4$

Answers (1)

best_answer

Energy emitted due to transition of electron -

$\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )$

$\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )$

- wherein

$R= R hydberg\: constant$

$n_{i}= initial state \\n_{f}= final \: state$

Total energy in initial excited state = $\frac{1240}{108.5}+\frac{1240}{30.4}$

= $52.22eV$

$52.22eV=13.6\times4\left[1-\frac{1}{n^{2}} \right ]$

$0.96=1-\frac{1}{n^{2}}$

$\Rightarrow \frac{1}{n^{2}}=0.04$

$\Rightarrow n^{2}=\frac{1}{0.04}=25$

$\Rightarrow n=5$

Option 1)

$n=5$

Option 2)

$n=6$

Option 3)

$n=7$

Option 4)

$n=4$

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