If kinetic energy of a proton is increased nine times, the wavelength of the de-Broglie wave associated with it would become

12mv2=K.E.  ⇒  12m2v2=m.K.E m2v2=2m .K.E λ=hmvsubstitute the value of mv mv=2m.K.E λ=h2m.K.E  ∴  λ∝1K.E......(1)From equation (1) when K.E. of the electron increased 9 times. The de – Broglie wavelength decreased by 13times.λ∝19=13