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 $E = {{1240\,eV} \over {\lambda (in\,nm)}}$) :

Question

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 $E = {{1240\,eV} \over {\lambda (in\,nm)}}$) :

n = 4

n = 7

n = 5

n = 6

Solution

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