The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearly

Explanation

if the kinetic energy is much less than the rest energy , we can use classical `KE=1/2mv^2` For an electron, `m_0 c^2`=0.511 MeV. We then apply conservation of energy : the KE acquired by the electron equals its loss in PE. After solving for v, we use `lambda=h/"mv" ` to that the de Broglie wavelength .
The gain in kinetic energy will equal the loss in potential energy (`DeltaPE`=eV-0) : KE=eV.
so KE=100 eV. The ratio `"KE"/(m_0c^2)="100 eV"/((0.511xx10^6 eV))approx 10^(-4)` , so relativity is not needed. Thus `1/2mv^2=eV` and `v=sqrt((2 eV)/m)`
`sqrt(((2)(1.6xx10^(-19)C)(100 V))/((9.1xx10^(-31) kg)))`
`=5.9xx10^6` m/s
Then `lambda=h/"mv"`
`=((6.63xx10^(-34) J.s))/((9.1xx10^(-31) kg)(5.9xx10^6 m//s))`
`=1.2xx10^(-10)` m or 0.12 nm