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Given ionisation potential of H atom is 13.6eV.Find the frequency and wavelength of Hβ line of Balmer series.
In Balmer series Hβ line transition is from n1=4 to n2=2
dE = 13.6*(n1-2 -n2-2)
= 13.6*3/16 = 2.55 eV
hc/(lamda) = 2.55 eV
lamda = hc / (2.55*1.6*10-19 )
wavelength = 486.12 nm
frequency = 6.17*1014 Hz
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Regards
Ramesh
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