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Calculate the de- Broglie wavelength of an electron having a kinetic energy of 1000eV.​

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Answer: λₑ = 123pm; λₚ = 286pm

Step-by-step explanation:

λ= h/√2mE where E = Kinetic Energy.

λₑ = h/√2mₑE = 6.626 X 10^-33/√2 X 9.1 X 10⁻31 X 100 X 1.6 X 10^-19 = 123pm

λₚ = h/√2mₚE = 6.626 X 10^-33/√2 X 1.6 X 10⁻27 X 100 X 1.6 X 10^-19 = 286pm

Hope this helps!!!

User Hassan Hayat
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