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  4. 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 initial excited state is (for photon of wavelength λ, energy E =(1240eV/λ(in nm))

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Q. 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 initial excited state is (for photon of wavelength $\lambda$, energy $E =\frac{1240eV}{\lambda\left(in\, nm\right)}$

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Solution:

$E = E_{1} +E_{2}$
$13.6 \frac{z^{2}}{n^{2}} = \frac{1240}{\lambda_{1}} +\frac{1240}{\lambda_{2}}$
or $\frac{13.6\left(2\right)^{2}}{n^{2}} = 1240$
$\left(\frac{1}{108.5} +\frac{1}{30.4}\right)\times\frac{1}{10^{-9}}$
On solving, $n = 5$