Question

For an electron accelerated from rest through a potential V, the de Broglie wavelength associated will be:

Options

• `1.772/sqrt"V" "nm"`

• `1.227/sqrt"V" μ"m"`

• `1.227/sqrt"V" "nm"`

• `1.772/sqrt"V" μ"m"`

Answer 1

`1.227/sqrt"V" "nm"`

Explanation:

λ = `1.227/sqrt"V" "nm"`, as λ = `"h"/"p" = "h"/sqrt(2"meV")`

Substituting the numerical values of h, m and e we get the result.