Question

Determine the de Broglie wavelength of an electron with a kinetic energy of 1.00 eV

Answer 1

Answer:

For an electron with KE = 1 eV and rest mass energy 0.511 MeV, the associated DeBroglie wavelength is 1.23 nm, about a thousand times smaller than a 1 eV photon

Answer 2

Explanation:

we get v=5.93m/s by calculation(v=2KEm)^1/2

then putting in de-broglie formula lamda=h/mv

$6.626 \times {10}^{ - 34} {js}^{ - 1}$

this is h

and mv= 9.1×10^-31×5.93

now put this in formula and get the answer.

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