What is the de Broglie wavelength associated with an electron, accelerated through a potential differnece of 100 volts?

Explanation

Accelerating potential `V = 100 V`. The de Broglie wavelength `lambda` is
`lambda=h//p=(1.227)/(sqrt(V))=nm`
`lambda=(1.227)/(sqrt(100))nm=0.123nm`
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.