Answer
a) The final wavelength is half of the initial wavelength.
b) $\lambda_{final}=\frac{\lambda_{inital}}{\sqrt2}$
Work Step by Step
As we know that
$\lambda=\frac{h}{p}$
$\implies p\times \lambda=h=constant$
$\implies p_{initial}\times \lambda_{initial}=p_{final}\times \lambda_{final}$
according to given condition $p_{final}=2p_{initial}$
$\implies p_{initial}\times \lambda_{initial}=2p_{initial}\times \lambda_{final}$
$\implies \lambda_{initial}=2\times \lambda_{final}$
$\lambda_{final}=\frac{\lambda_{inital}}{2}$
Thus, the final wavelength is half of the initial wavelength.
(b) We know that $K.E\propto (momentum)^2$. Thus, when kinetic energy is doubled then final momentum will be $\sqrt 2$ of initial momentum
As $\lambda=\frac{h}{p}$
$\implies p\times \lambda=h=constant$
$\implies p_{initial}\times \lambda_{initial}=p_{final}\times \lambda_{final}$
according to given condition $p_{final}=\sqrt 2p_{initial}$
$\implies p_{initial}\times \lambda_{initial}=\sqrt2p_{initial}\times \lambda_{final}$
$\implies \lambda_{initial}=\sqrt2\times \lambda_{final}$
$\lambda_{final}=\frac{\lambda_{inital}}{\sqrt2}$
