If the shortest wavelength of H-atom in Lyman series is 'X' then the longest wavelength in Balmer
series of He+ is
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longest wavelength(λl) in Balmer series means 3→ 2 transition
and shortest wavelengtha(λs) in Lyman series means infinte orbit → 1 transition which is equal to x
According to question λs=Rh[121−i21]=Rh=x ( i= infinite )
λl=Rh[221−321]z2=Rh[365]4=Rh95=95x
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