1. Tardigrade
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  3. Physics
  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 = (1240 eV/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 $E = \frac{1240 eV}{\lambda (in \, nm)}$ ;

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Solution:

$\frac{1}{\lambda} = R \left(\frac{1}{m^{2}}- \frac{1}{n ^{2}}\right)z^{2} $
$ \frac{1}{1085} =R \left(\frac{1}{m^{2}} - \frac{1}{n^{2}}\right)2^{2} $
$ \frac{1}{304} =R \left(\frac{1}{1^{2}} - \frac{1}{m^{2}}\right)2^{2} $
$ \therefore m = 2$
$ \therefore n = 5 $