Answer
(a) decrease
(b)$\lambda= \frac{1.23}{\sqrt K}$nm$\sqrt{eV}$
Work Step by Step
(a) We know that $\lambda\propto \frac{1}{K.E}$. This shows that the de Broglie wavelength is inversely proportional to the kinetic energy. Thus, if the kinetic energy is increased, then the de Broglie wavelength of the particle decreases.
(b) We know that
$\lambda=\frac{h}{\sqrt {2mK}}$
We plug in the known values to obtain:
$\lambda=\frac{6.63\times 10^{-34}J.s}{\sqrt{(2)(9.11\times 10^{-31})K(eV)}}$
$\lambda=\frac{6.63\times 10^{-34}J.s}{\sqrt{18.22\times 10^{-31}Kg}\sqrt{K(1.6\times 10^{-19}J)}}$
$\lambda= \frac{1.23}{\sqrt K}$nm$\sqrt{eV}$