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Given ionisation potential of H atom is 13.6eV.Find the frequency and wavelength of Hβ line of Balmer series.

ANKUR GUPTA , 15 Years ago
Grade 12
anser 1 Answers
Ramesh V

Last Activity: 15 Years ago

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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