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Q: 11.13 (c)  What is the de Broglie wavelength of an electron with kinetic energy of 120\hspace{1mm}eV

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De Broglie wavelength is given by

\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{5.911\times 10^{-24}}\\ \lambda =1.12\times 10^{-10}\ m\\

The de Broglie wavelength associated with the electron is 0.112 nm

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