An excited He^(+) ion emits two photons in succession, with wavelength 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 lamda, energy E=(1240eV)/(lamda("in nm"))

An excited He^(+) ion emits two photons in succession, with wavelength

1.	An excited He^(+) ion emits two photons in succession, with wavelength 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 lamda, energy E=(1240eV)/(lamda("in nm"))
Answer» N= 6 n = 5 n = 7 n = 4 Answer :B

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An excited He^(+) ion emits two photons in succession, with wavelength 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 lamda, energy E=(1240eV)/(lamda("in nm"))

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