if de Broglie wavelength becomes halved momentum becomes doubled and kinetic energy becomes 4 times  and if incident wavelength decreases by 60%  it means it becomes 40%  

$$\begin{gathered} \lambda becomes \frac{4}{10}\lambda frequency becomes \frac{10}{4}\nu \\ Eienstein equation h\nu =w+k----\left(1\right) \\ second time h \frac{10}{4}\nu = w+4k----\left(2\right) \\ solving w=0.5 h\nu here h\nu = 2ev \end{gathered}$$