1.

What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 volt ?

Answer»

`12.27 Å`
`1.227Å`
`0.1227Å`
`0.001227Å`

SOLUTION :`LAMBDA = (12.27)/(SQRT(V)) = (12.27)/(sqrt(100)) = 1.227 A^(@)`


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