The wavelength of one line in the visible region o fthe atmoic spectrum of hydrogen is `6.5xx10^(-7)m`. This radiation is emitted when electron in a hydrogen atom goes from a high enegry state to a lower energy state. Calculate the difference in enegry between the two states.
Strategy: According to the Bohr frequency rule, if the electron jumps from one orbit whose quantum number is `n_(1)` to a second orbit whose quantum number is `n_(f)`, the difference in enegry `(DeltaE)` is related to the frequency `(v)` of the radiation which in turn is related to the wavelength `(lambda)`:
? `DeltaE = hv` and `v = (c)/(lambda)`
Thus, `DeltaE = (hc)/(lambda)`

Explanation

Insert the values of the constants and wavelengthin the above expression to obtain
`DeltaE = ((6.6xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/((6.5xx10^(-7)m))`