Answer
(a) $\lambda = 1.23\times 10^{-9}~m$
(b) $\lambda = 3.88\times 10^{-11}~m$
Work Step by Step
(a) We can write an expression for the momentum:
$p = \sqrt{2mE}$
We can find the de Broglie wavelength:
$\lambda = \frac{h}{p}$
$\lambda = \frac{h}{\sqrt{2mE}}$
$\lambda = \frac{6.626\times 10^{-34}~J~s}{\sqrt{(2)(9.1\times 10^{-31}~kg)(1.0~eV)(1.6\times 10^{-19}~J/eV)}}$
$\lambda = 1.23\times 10^{-9}~m$
(b) We can write an expression for the momentum:
$p = \sqrt{2mE}$
We can find the de Broglie wavelength:
$\lambda = \frac{h}{p}$
$\lambda = \frac{h}{\sqrt{2mE}}$
$\lambda = \frac{6.626\times 10^{-34}~J~s}{\sqrt{(2)(9.1\times 10^{-31}~kg)(1000~eV)(1.6\times 10^{-19}~J/eV)}}$
$\lambda = 3.88\times 10^{-11}~m$
