What is the de Broglie wave lenght of an electron, which is accelerated from the rest, through a potential difference of 100 V?

Explanation

Accelerating potential V = 100 V.
The de Broglie wavelength, `lambda=h//p=(1.227)/(sqrt(V))nm=(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.