De broglie wavelength is given by :

lambda=h/mv, and since kinetic energy=1/2(mv^2)=qV where q is the charge, and V the potential difference

Therefore lambda=h/((2mqV)^0.5))

Putting values,

h= 6.62607004 × 10 ^-34 m ² kg / s

m=9.10938356 × 10 ^-31 kilograms

q=1.60217662 × 10 ^-19 coulombs

V= 100 volts

We get : lambda=1.2293 Å