Questions
The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometre is
detailed solution
Correct option is B
λ=h2mE⇒E=h22mλ2 =(6.6×10-34)22×9.1×10-31×(0.3×10-9)2 =2.65×10-18J=16.8 eV
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