de-Broglie wavelength of an electron is given by
$\lambda =\frac{h}{m v}=\frac{h}{\sqrt{2 m e V}}=\frac{12.27}{\sqrt{V}}\mathring{A}$
(h=Planck's constant)
where $m=$ mass of electron
$e=$ electronic charge and
$V=$ potential difference with which electron is accelerated
$\lambda =\frac{12.27}{\sqrt{100}} \mathring{A}=\frac{12.27}{10} \mathring{A}$
$=1.227\, \mathring{A}$