We see form the previous problem that the electron gains the maximum `K.E.` when the photon is scattered backwards `theta = 180^(@)`. Then
`omega_(0) = omega_(0) = (mc^(2)//cancel h)/(sqrt(1+(2mc^(2))/(T_(max)) )-1)`
Hence `lambda_(0) = (2pic)/(omega_(0)) = (2picancel h)/(mc) [sqrt(1+(2mc^(2))/(T_(max)))-1]`
Substituting the values we get `= lambda_(0) = 3.695 pm`.