Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.
λ=h/√(2mE)=12.27/√V Å=12.27/√120≈1.12 Å per de Broglie formula.
Ref: NCERT Physics Class 12 Chapter 11 Dual Nature of Radiation and Matter
#120 eV
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