In hydrogen atom, an electron jumps from 3d orbit to the 2nd orbit. Calculate the wavelength of the radiatoin emitted. (h = 6.63 xx 10^(-34) J sec)

In hydrogen atom, an electron jumps from 3d orbit to the 2nd orbit. Ca

1.	In hydrogen atom, an electron jumps from 3d orbit to the 2nd orbit. Calculate the wavelength of the radiatoin emitted. (h = 6.63 xx 10^(-34) J sec)
Answer» Solution :`E_(N) = - (21.8 xx 10^(-19))/(n^(2)) J "atom"^(-1)` `Delta E = E_(3) - E_(2) = 21.8 xx 10^(-19) ((1)/(2^(2)) - (1)/(3^(2))) = 3.03 xx 10^(-19) J` `Delta E = hv = H (c)/(lamda)` `lamda = (HC)/(Delta E) = ((6.63 xx 10^(-34) Js) (3 xx 10^(8) MS^(-1)))/((3.03 xx 10^(-19) J)) = 6.564 xx 10^(-7) m = 6564 xx 10^(-10) m =Å`