Answer
(a) $3.885\times 10^{-11}$ $m$
(b) $1.243\times 10^{-9}$ $m$
(c) $9.069\times 10^{-13}$ $m$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$, ......................$(1)$
where $h$ is the Planck's constant and $p$ is the momentum of the moving particle..
(a) For electron, the relation between momentum $p$ and energy $E$ is
$E=\frac{p^2}{2m}$ , where $m$ is the mass of the electron.
Thus, $p=\sqrt {2mE}$ ......................$(2)$
Substituting eq. $2$ in eq. $1$, we get
$\lambda=\frac{h}{\sqrt {2mE}}$ ......................$(3)$
Now for an electron having energy $1.00$ $keV$
$\lambda=\frac{6.63\times 10^{-34}}{\sqrt {2\times9.1\times 10^{-31}\times 1\times 10^3\times 1.6\times 10^{-19}}}$
or, $\lambda=3.885\times 10^{-11}$ $m$
(b) For photon, the relation between momentum $p$ and energy $E$ is
$E=pc$
Thus, $p=\frac{E}{c}$ ......................$(4)$
Substituting eq. $4$ in eq. $1$, we get
$\lambda=\frac{hc}{E}$
Now for a photon having energy $1.00$ $keV$
$\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{1\times 10^3\times 1.6\times 10^{-19}}$ $m$
or, $\lambda=1.243\times 10^{-9}$ $m$
(c) As neutron has finite mass, we can use eq. $3$ to calculate the de Broglie wavelength ($\lambda$)
Now for a neutron having energy $1.00$ $keV$
$\lambda=\frac{6.63\times 10^{-34}}{\sqrt {2\times 1.67\times 10^{-27}\times 1\times 10^3\times 1.6\times 10^{-19}}}$
or, $\lambda=9.069\times 10^{-13}$ $m$
